For the following reaction, 33.5 grams of **iron** are allowed to react with 18.8 grams of **oxygen gas**.**iron(s) + oxygen(g) ⟶ iron(III) oxide(s)**1. What is the maximum mass of **iron(III) oxide** that can be formed? [______] grams2. What is the **FORMULA** for the limiting reagent? [______]3. What mass of the excess reagent remains after the reaction is complete? [______] grams For the following reaction, 7.70 grams of sodium are allowed to react with 10.4 grams of water.\[ \text{sodium(s) + water(l)} \rightarrow \text{sodium hydroxide(aq) + hydrogen(g)} \]- **What is the maximum amount of sodium hydroxide that can be formed?** [ ] grams- **What is the FORMULA for the limiting reagent?** [ ]- **What amount of the excess reagent remains after the reaction is complete?** [ ] gramsIn this chemical reaction, solid sodium reacts with liquid water to produce aqueous sodium hydroxide and gaseous hydrogen. The task involves calculating the maximum amount of sodium hydroxide that can be produced, identifying the limiting reagent, and determining the leftover amount of the excess reagent after the reaction has concluded.

Hello. Since more than one question has been asked, the first question shall only be answered in this case. If the answer of the remaining question is required, please specify the questions and submit them. Thanks.Given mass of iron = 33.5 g Molar mass of iron = 55.85 g/mol Number of moles of iron = 33.5/55.85 = 0.5998 mol Given mass of oxygen gas = 18 g Molar mass of oxygen gas = 32 g/mol Number of moles of oxygen gas = 18/32 = 0.5625 mol The balanced chemical equation can be written as follows: 4Fe + 3O2 → 2Fe2O3Iron is the limiting reagent in the chemical reaction. The formula of the limiting reagent is Fe. Number of moles of iron (III) oxide formed by 4 moles of iron = 2 mol Number of moles of iron (III) oxide formed by 1 mol of iron = 0.5 mol Number of moles of iron (III) oxide formed by 0.5998 mol of iron = 0.5998 × 0.5 = 0.2999 mol Mass of 1 mole of iron (III) oxide = 55.85 + 55.85 + 48 = 159.7 g Mass of 0.2999 mol of iron (III) oxide = 159.7 × 0.2999 = 47.89 g Mass of oxygen gas that reacts with 223.4 g of Fe = 96 g Mass of oxygen gas that reacts with 1 g of iron = 96/223.4 = 0.4297 g Mass of oxygen gas that reacts with 33.499 g of iron = 33.499 × 0.4297 = 14.395 g Mass of oxygen gas (excess reagent) that remains after the reaction = 18 – 14.395 = 3.605 g

Answered: eaction, 33 grams худеп gas iron(s) +…

Science

Chemistry

eaction, 33 grams худеп gas iron(s) + oxygen(g - iron(III) oxide(s) What is the maximum mass of iron(III) oxide that can be formed? | grams What is the FORMULA for the limiting reagent? |

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

Similar questions

For the following reaction, 19.3 grams of sulfur dioxide are allowed to react with 10.3 grams of water. sulfur dioxide(g)+ water(1) - sulfurous acid (H,SO3)(g) What is the maximum mass of sulfurous acid (H,SO3) that can be formed? | grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams
Before arc welding was developed, a displacement reactioninvolving aluminum and iron(III) oxide was commonly used toproduce molten iron (the thermite process). This reaction wasused, for example, to connect sections of iron railroad track. Cal-culate the mass of molten iron produced when 1.50 kg of alu-minum reacts with 25.0 mol of iron(III) oxide.
15)
3. If you do the reaction in the lab with 32.0 grams of NaCl and the perfect stoichiometric amount of H,SO, and measure that you only make 32.42 grams (your actual yield) of Na, SO.. what is your percent yield of Na,SO,? (Hint: Your answer to #1g is yourleheoretical yield.) % Yield = Actual Yield x 100 Theoretical Yield
Gaseous ethane (CH, CH;) reacts with gaseous oxygen gas (0,) to produce gaseous carbon dioxide (Co,) and gaseous water (H, O). What is the theoretical yield of carbon dioxide formed from the reaction of 0.60 g of ethane and 1.9 g of oxygen gas? Be sure your answer has the correct number of significant digits in it. g
The illustration to the left represents a mixture of nitrogen (blue) and oxygen (red) molecules. If the molecules in the above illustration react to form N₂O4 according to the equation N₂ + 20₂→→→→→→→ N₂O4, the limiting reagent is the number of N₂04 molecules formed is and the number of molecules in excess is
A east.cengagenow.com Chemistry I E OWLV2 | Online teaching and learning resource from Cengage Learning According to the following reaction, how many gram [References) Use the References to access important values if needed for this question. According to the following reaction, how many grams of water will be formed upon the complete reaction of 25.8 grams of oxygen gas with excess hydrogen sulfide? hydrogen sulfide (g) + oxygen (g) water (I) + sulfur dioxide (g) grams water Submit Answer Try Another Version 5 Item attempts remaining Cengage Learning Cengage Technical Support
For the following reaction, 6.48 grams of methane (CH,) are allowed to react with 26.7 grams of carbon tetrachloride . methane (CH4)(g) + carbon tetrachloride(g) dichloromethane (CH2Cl,)(g) What is the maximum mass of dichloromethane (CH,Cl,) that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams
Seven please
This is stoichiometry And it’s for % yield
< [Review Topics] [References) Use the References to access important values if needed for this question. For the following reaction, 0.577 moles of methane (CH4) are mixed with 0.399 moles of carbon tet methane (CH4) (g) + carbon tetrachloride (g) → dichloromethane (CH₂ Cl₂) (g) What is the formula for the limiting reactant? What is the maximum amount of dichloromethane (CH₂ Cl₂) that can be produced? Amount = mol Submit Answer Retry Entire Group 4 more group attempts remaining
For the following reaction, 3.76 grams of hydrogen gas are allowed to react with 40.5 grams of iodine . hydrogen(g) + iodine(s) → hydrogen iodide(g) What is the maximum mass of hydrogen iodide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 7 more group attempts remaining