What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 62.5 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction?Na₃PO₄(aq) + Cr(NO₃)₃(aq) → CrPO₄(s) + 3 NaNO₃(aq) **Problem Statement:**What mass of precipitate (in grams) is formed when 20.5 mL of 0.800 M Co(NO₃)₂ reacts with 16.5 mL of 0.800 M NaOH in the following chemical reaction?**Chemical Reaction:**Co(NO₃)₂(aq) + 2 NaOH(aq) ⟶ Co(OH)₂(s) + 2 NaNO₃(aq)---**Explanation:**This problem involves a precipitation reaction where cobalt(II) nitrate reacts with sodium hydroxide to form cobalt(II) hydroxide as a solid precipitate and sodium nitrate in solution. To find the mass of the precipitate formed, follow these steps:1. **Calculate Moles of Reactants:** - **Co(NO₃)₂:** Use the formula \( \text{Moles} = \text{Volume (L)} \times \text{Molarity (M)} \). - **NaOH:** Similarly, calculate using the same formula.2. **Determine Limiting Reagent:** - Compare the mole ratio from the balanced equation to find which reactant is the limiting reagent.3. **Calculate Moles of Precipitate:** - Based on the limiting reagent, use stoichiometry to find moles of Co(OH)₂ formed.4. **Find Mass of Precipitate:** - Convert moles of Co(OH)₂ to mass using its molar mass.This process outlines how to solve for the mass of precipitate using stoichiometry and reactant quantities in a chemical reaction.

Since, Molarity is the number of moles of solute dissolve in one litre of the solution. Thus,

What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na PO reacts with 62.5 mL of 0.200 M Cr(NO) in the following chemical reaction? Na₂PO₂(aq) + Cr(NO³)²(aq) → CrPO₂(s) + 3 NaNO₂(aq)

What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na PO reacts with 62.5 mL of 0.200 M Cr(NO) in the following chemical reaction? Na₂PO₂(aq) + Cr(NO³)²(aq) → CrPO₂(s) + 3 NaNO₂(aq)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Q: 0.7532 g of copper turnings (molar mass: 63.546 g∙mol–1) are reacted with excess nitric acid.…

A: Disclaimer: In order to obtain copper sulfate the acid used in excess should be sulfuric acid and…

Q: What quantity in moles of precipitate will be formed when 56.2 mL of 0.250 M AgNO3 is reacted with…

A: The volume of AgNO3 is .The molarity of AgNO3 is .The given balanced chemical equation is .To find…

Q: If 33.0 g of NaOH is added to 0.650 L of 1.00 M Cu(NO₃)₂, how many grams of Cu(OH)₂ will be formed…

A: First of all, converting everything into the number of moles will help us to find the limiting…

Q: 100.0 mL of an aqueous solution containing 0.250 M KOH(aq) is mixed with 100.0 mL of an aqueous…

A: First of all, we need to write down the balanced chemical. The molar mass of Fe(OH)3 = 106.87g/mol

Q: What quantity in moles of precipitate will be formed when 45.1 mL of 0.250 M AgNO₃ is reacted with…

Q: How many moles of precipitate will be formed when 85.3 mL of 0.350 M AgNO₃ is reacted with excess…

A: Given data,Molarity of AgNO3=0.350MVolume of AgNO3=85.3mL=0.0853L

Q: What quantity in moles of precipitate will be formed when 272.5 mL of 0.400 M NaI is reacted with…

Q: How many moles of precipitate will be formed when 70.0 mL of 2.00 M NaBr is reacted with 300.0 mL of…

A: The balanced chemical reaction:2NaBr(aq) + Pb(NO3)2(aq) → PbBr2(s) + 2NaNO3(aq)The volume of 2.00 M…

Q: If 22.0 g of NaOH is added to 0.700 L of 1.00 M Co(NO₃)₂, how many grams of Co(OH)₂ will be formed…

A: The balanced reaction taking place is given as, => 2 NaOH (aq) + Co(NO3)2 (aq) → Co(OH)2 (s) + 2…

Q: What mass of precipitate (in g) is formed when 20.5 mL of 0.300 M Ni(NO₃)₂ reacts with 33.0 mL of…

Q: What quantity in moles of precipitate will be formed when 77.1 mL of 0.150 M AgNO3 is reacted with…

Q: What quantity in moles of precipitate will be formed when 73.7 mL of 0.150 M AgNO₃ is reacted with…

Q: you begin with 78.2 mL of 0.725 M KCl(aq) solution and mix it with 67 mL of 0.721 M Pb(NO3)2(aq) ,…

A: Given-> Volume of KCl= 78.2 ml Molarity of KCl = 0.725 M Volume of Pb(NO3)2(aq) = 67 ml Molarity…

Q: If 19.5 g of NaOH is added to 0.800 L of 1.00 M Zn(NO₃)₂, how many grams of Zn(OH)₂ will be formed…

A: Given, Mass of NaOH = 19.5 g Molarity of Zn(NO₃)₂ = 1.00 M Volume of Zn(NO₃)₂…

Q: According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M…

A: Moles of LiOH used =Molarity of LiOH×Volume of LiOH= 0.227 mol/L×175.0×10-3L=0.0397 mol

Q: What quantity in moles of precipitate will be formed when 168.0 mL of 0.350 M NaCl is reacted with…

Q: How many moles of precipitate will be formed

A: Molarity = number of moles/volume of solution (in L) Number of moles = molarity × volume (in L)

Q: What volume of 0.250 M Na3PO4 is required to precipitate all the lead(II) ions from 190.0 mL of…

A: Molarity is defined as number of mole of solute divide by volume of solution in liter

Q: What mass of precipitate (in g) is formed when 20.5 mL of 0.700 M Zn(NO₃)₂ reacts with 23.0 mL of…

A: Number of moles of substance=Concentration of substance x Volume of substanceFor…

Q: How many moles of precipitate will be formed when 80.0 mL of 2.00 M NaI is reacted with 100.0 mL of…

A: 2 NaI (aq) + Pb(NO₃)₂ (aq) → PbI₂ (s) + 2 NaNO₃ (aq) To determining limiting reactant ; Firstly we…

Q: If 29.0 g of NaOH is added to 0.700 L of 1.00 M Co(NO₃)₂, how many grams of Co(OH)₂ will be formed…

A: Since , no of moles of CoNO32= Concentration of solution x…

Q: What quantity in moles of precipitate will be formed when 70.0 mL of 0.150 M AgNO₃ is reacted with…

A: Consider the given reaction is as follows; 2 AgNO3 (aq) + CaCl2 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq)…

Q: How many moles of precipitate will be formed when 30.0 mL of 0.450 M AgNO₃ is reacted with excess…

Q: What mass of precipitate (in g) is formed when 20.5 mL of 0.700 M Zn(NO₃)₂ reacts with 31.0 mL of…

A: ZnNO32+2 NaOH→ZnOH2 s+2 NaNO3 aqStength of ZnNO32=20.5x0.700 M1000=0.0143Stength of NaOH=310x0.700…

Q: How many moles of precipitate will be formed when 70.2 mL of 0.150 M AgNO₃ is reacted with excess…

Q: What mass of precipitate (in g) is formed when 20.5 mL of 0.300 M Ni(NO3)2 reacts with 40.0 mL of…

Q: what mass in grams of Zn(OH)₂ will be formed

A: Number of moles = mass/molar mass Molarity = moles/volume(in L) Moles = molarity × volume(in L)…

Q: If 25.5 g of NaOH is added to 0.800 L of 1.00 M Zn(NO₃)₂, what mass in grams of Zn(OH)₂ will be…

Q: What mass of precipitate (in g) is formed when 20.5 mL of 0.300 M Ni(NO₃)₂ reacts with 22.5 mL of…

A: According law of mass action, the same amount of mass involved in the reaction is formed as a…

Q: How many moles of precipitate will be formed when 112.0 mL of 0.350 M NaCl is reacted with excess…

A: Given reaction: 2 NaCl (aq) + Pb(NO₃)₂ (aq) → PbCl₂ (s) + 2 NaNO₃ (aq) volume of 0.350 M NaCl…

Q: What mass of precipitate (in g) is formed when 20.5 mL of 0.500 M Cu(NO₃)₂ reacts with 34.5 mL of…

A: First calculate moles for both solutions. Then using mol to mol ratio, find out moles of precipitate…

Q: mass

Q: What quantity in moles of precipitate will be formed when 45.6 mL of 0.550 M AgNO3 is reacted with…

Q: How many moles of precipitate will be formed when 100 mL of 2.00 M NaBr is reacted with 70.0 mL of…

A: The mole of precipitate will be formed has to be determined

Q: How many moles of precipitate will be formed when 74.2 mL of 0.150 M AgNO₃ is reacted with excess…

Q: If 28.0 g of NaOH is added to 0.550 L of 1.00 M Ni(NO₃)₂, how many grams of Ni(OH)₂ will be formed…

A: In the reaction ; 2 NaOH(aq) + Ni(NO₃)₂(aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq) So, we can conclude that 2…

Q: If 33.0 g of NaOH is added to 0.550 L of 1.00 M Ni(NO₂), how many grams of Ni(OH)₂ will be formed in…

A: we have to calculate the mass of precipitate formed in the given reaction

Q: How many moles of precipitate will be formed when 69.3 mL of 0.350 M AgNO₃ is reacted with excess…

A: Given the chemical reaction as follows,2 AgNO3(aq) + MgI2(aq) → 2 AgI(s) + Mg(NO3)2(aq)we are asked…

Q: You have 0.685 g of an unknown acid, H₂A, which reacts with NaOH according to the balanced equation…

Q: If 28.5 g of NaOH is added to 0.800 L of 1.00 M Zn(NO₃)₂, what mass in grams of Zn(OH)₂ will be…

A: Given data mass of NaOH = 28.5 g volume of Zn(NO3)2 = 0.800 L molarity of Zn(NO3)2 = 1.00 M Reaction…

Q: What quantity in moles of precipitate will be formed when 75.0 mL of 2.00 M NaCl is reacted with…

Q: A titration experiment required 16.64 mL of NaOH, which had a concentration of 0.135 M, to reach the…

Q: If 32.5 g of NaOH is added to 0.550 L of 1.00 M Ni(NO₃)₂, how many grams of Ni(OH)₂ will be formed…

A: The reaction considered is, 2 NaOH(aq)+Ni(NO3)2(aq)→Ni(OH)2 (s)+ 2 NaNO3(aq)

Q: If 16.5 g of NaOH is added to 0.550 L of 1.00 M Ni(NO₃)₂, how many grams of Ni(OH)₂ will be formed…

A: The balanced reaction given is 2 NaOH(aq) + Ni(NO₃)₂(aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq) Given : Mass…

Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 62.5 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction?

Na₃PO₄(aq) + Cr(NO₃)₃(aq) → CrPO₄(s) + 3 NaNO₃(aq)

**Problem Statement:**

What mass of precipitate (in grams) is formed when 20.5 mL of 0.800 M Co(NO₃)₂ reacts with 16.5 mL of 0.800 M NaOH in the following chemical reaction?

**Chemical Reaction:**

Co(NO₃)₂(aq) + 2 NaOH(aq) ⟶ Co(OH)₂(s) + 2 NaNO₃(aq)

---

**Explanation:**

This problem involves a precipitation reaction where cobalt(II) nitrate reacts with sodium hydroxide to form cobalt(II) hydroxide as a solid precipitate and sodium nitrate in solution. To find the mass of the precipitate formed, follow these steps:

1. **Calculate Moles of Reactants:**
- **Co(NO₃)₂:** Use the formula \( \text{Moles} = \text{Volume (L)} \times \text{Molarity (M)} \).
- **NaOH:** Similarly, calculate using the same formula.

2. **Determine Limiting Reagent:**
- Compare the mole ratio from the balanced equation to find which reactant is the limiting reagent.

3. **Calculate Moles of Precipitate:**
- Based on the limiting reagent, use stoichiometry to find moles of Co(OH)₂ formed.

4. **Find Mass of Precipitate:**
- Convert moles of Co(OH)₂ to mass using its molar mass.

This process outlines how to solve for the mass of precipitate using stoichiometry and reactant quantities in a chemical reaction.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Step 5

Step by step

Solved in 5 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

How many grams of PbBr2 will precipitate when excess NiBr2 solution is added to 47.0 mL of 0.602 M Pb(NO3)2 solution? Pb(NO3)2 (aq) + NiBr2 (aq) → PbBr2 (s) + Ni(NO3)2 (aq) g
What quantity in moles of precipitate will be formed when 100.0 mL of 2.00 M NaI is reacted with 80.0 mL of 0.900 M Pb(NO₃)₂ in the following chemical reaction?2 NaI(aq) + Pb(NO₃)₂(aq) → PbI₂(s) + 2 NaNO₃(aq)
How many moles of precipitate will be formed when 47.5 mL of 0.250 M AgNO₃ is reacted with excess CaBr₂ in the following chemical reaction? 2 AgNO₃(aq) + CaBr₂(aq) → 2 AgBr(s) + Ca(NO₃)₂(aq)
What mass of precipitate (in g) is formed when 20.5 mL of 0.300 M Ni(NO₃)₂ reacts with 25.5 mL of 0.300 M NaOH in the following chemical reaction? Ni(NO₃)₂(aq) + 2 NaOH(aq) → Ni(OH)₂(s) + 2 NaNO₃(aq)
How many moles of iron(II) hydroxide are produced when 750 mL of a 1.8 M hydroxide ion solution are mixed with an excess of iron(II) ions? The net ionic equation is Fe2+ (aq) + 2 OH¯ (aq) → Fe(OH)2 (s) 1.4 mol Fe(OH)2 0.68 mol Fe(OH)2 2.4 mol Fe(OH)2 1.2 mol Fe(OH)2
What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Li₃PO₄ reacts with 50.5 mL of 0.200 M CrCl₃ in the following chemical reaction? Li₃PO₄(aq) + CrCr₃(aq) → CrPO₄(s) + 3 LiCl(aq)
How many moles of precipitate will be formed when 71.7 mL of 0.150 M AgNO₃ is reacted with excess CaCl₂ in the following chemical reaction? 2 AgNO₃(aq) + CaCl₂(aq) → 2 AgCl(s) + Ca(NO₃)₂(aq)
Consider the following reaction. MgCl2(aq) + 2 NaOH(aq) → Mg(OH)2(s) + 2 NaCl(aq) A 185.0 mL solution of 0.395 M MgCl2 reacts with a 37.62 mL solution of 0.811 M NaOH to produce Mg(OH)2 and NaCl. Identify the limiting reactant. ○ Mg(OH)2 NaOH MgCl2 NaCl Caclulate the mass of Mg(OH)2 that can be produced. mass of Mg(OH)2: The actual mass of Mg(OH)2 isolated was 0.742 g. Calculate the percent yield of Mg(OH)2. percent yield: 6.0 g %
if 21.5 g of NaOH is added to 0.800 L of 1.00 M Zn(NO₃)₂, what mass in grams of Zn(OH)₂ will be formed in the following precipitation reaction? 2 NaOH(aq) + Zn(NO₃)₂(aq) → Zn(OH)₂ (s) + 2 NaNO₃ (aq)
What quantity in moles of precipitate will be formed when 87.6 mL of 0.150 M AgNO₃ is reacted with excess CaCl₂ in the following chemical reaction? 2 AgNO₃(aq) + CaCl₂(aq) → 2 AgCl(s) + Ca(NO₃)₂(aq)
What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Li₃PO₄ reacts with 56.0 mL of 0.200 M CrCl₃ in the following chemical reaction? Li₃PO₄(aq) + CrCr₃(aq) → CrPO₄(s) + 3 LiCl(aq)
You have 0.765 g of an unknown acid, H₂A, which reacts with NaOH according to the balanced equation H₂A(aq) + 2 NaOH(aq) → Na2 A(aq) + 2 H₂O(l) If 25.78 mL of 0.461 M NaOH is required to titrate the acid to the second equivalence point, what is the molar mass of the acid? Molar mass = g/mol