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

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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...

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