**Title: Chemistry Problem - Molarity and Stoichiometry****Problem Statement:**How many milliliters (mL) of 0.776 M HCl are needed to dissolve 6.95 grams of CaCO₃?**Chemical Reaction:**\[ 2\text{HCl}(aq) + \text{CaCO}_3(s) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) \]\[ \fbox{ } \text{ mL} \]---**Explanation:**This problem involves using the principles of molarity and stoichiometry to determine the volume of hydrochloric acid (HCl) needed to fully react with a given amount of calcium carbonate (CaCO₃). 1. **Identify the Chemical Equation:** The balanced chemical equation is provided, illustrating that 2 moles of hydrochloric acid react with 1 mole of calcium carbonate to produce calcium chloride, water, and carbon dioxide.2. **Calculate Moles of CaCO₃:** - First, determine the molar mass of CaCO₃: - Calcium (Ca): 40.08 g/mol - Carbon (C): 12.01 g/mol - Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol - Total molar mass of CaCO₃ = 40.08 + 12.01 + 48.00 = 100.09 g/mol - Calculate moles of CaCO₃: \[ \text{Moles of CaCO₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{6.95 \text{ g}}{100.09 \text{ g/mol}} \approx 0.0694 \text{ mol} \]3. **Use Stoichiometry to Find Moles of HCl Needed:** - According to the balanced equation, 2 moles of HCl are required for every 1 mole of CaCO₃. - Moles of HCl needed = 0.0694 mol CaCO₃ × 2 mol HCl / 1 mol CaCO₃ = 0.1388 mol HCl4. **Calculate Volume **Reaction Stoichiometry Calculation**In this exercise, we aim to calculate the number of milliliters of a 0.689 M NaOH solution required to precipitate all of the Al³⁺ ions present in 163 mL of a 0.784 M Al₂(SO₄)₃ solution.**Chemical Reaction:**The balanced chemical equation for the reaction is:\[ \text{Al}_2(\text{SO}_4)_3 (\text{aq}) + 6\text{NaOH} (\text{aq}) \rightarrow 2\text{Al}(\text{OH})_3 (\text{s}) + 3\text{Na}_2\text{SO}_4 (\text{aq}) \]**Calculation Steps:**1. **Determine Moles of Al₂(SO₄)₃:** - Molarity (M) of Al₂(SO₄)₃ = 0.784 M - Volume (V) of Al₂(SO₄)₃ = 163 mL = 0.163 L - Moles of Al₂(SO₄)₃ = Molarity × Volume = 0.784 M × 0.163 L2. **Relate Moles of Al₂(SO₄)₃ to Moles of NaOH:** - From the balanced equation, 1 mole of Al₂(SO₄)₃ reacts with 6 moles of NaOH.3. **Calculate Volume of NaOH Solution Required:** - Molarity (M) of NaOH = 0.689 M - Moles of NaOH = (Moles of Al₂(SO₄)₃) × 6 - Volume of NaOH = Moles of NaOH / Molarity of NaOHBy following these steps, we can find out the volume in milliliters of the NaOH solution needed.**Volume Calculation:**\[ \text{Volume} = \boxed{\text{ }} \text{mL} \]This computational step is crucial in helping students understand reaction stoichiometry, as well as the practical application of balancing chemical equations in real-world laboratory settings.

Since each question is different, we solve only first question according to the Bartleby guidelines.…

Answered: How many mL of 0.776 M HCI are needed…

How many mL of 0.776 M HCI are needed to dissolve 6.95 g of CaCO3? 2HCl(aq) + CaCO3(s) →→→ CaCl₂(aq) + H₂O(l) + CO₂(g) mL

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...

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