please answer only part e **Tutorial: Limiting Reactant Determination**This tutorial will guide you through the steps of determining which reactant is limiting based on a balanced chemical equation.a) **Step 1: Evaluate the question.** Which of the following correctly describes a limiting reactant?b) **Calculate the theoretical yield in moles of CO₂** from the complete combustion of 29.3 grams of CH₄ according to the balanced chemical equation: \[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow 1 \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \]c) **Calculate the theoretical yield in moles of CO₂** from the complete reaction of 81.7 grams of O₂ according to the balanced chemical equation: \[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow 1 \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \]d) Which of the following is the limiting reactant?e) **What is the mass in grams of the excess CH₄** remaining after the partial combustion in 81.7 grams O₂?

The chemical reaction is given as, CH4 + 2O2 --------> CO2 + 2H2O We need to find excess ofmethane when 29.3 g of methane react with 81.7 g of oxygen. This we can solve using the balanced chemical reaction. e) According to the balanced chemical reaction, 1 mole of methane required 2 moles of O2. Number of moles = given mass/molar mass n = w/M nCH4 = 29.3/16.04 = 1.83 moles nO2 = 81.7/32 = 2.55 moles So, 2.55 moles of O2 required 2.55/2 = 1.28 moles of methane. So the remaining number of moles of methane after the complete reaction is, n = 1.83 - 1.28 n = 0.55 moles So the mass of CH4 remaining = 0.55×16.04 = 8.82 g So the answer is 8.82 g of excess CH4 will remain after partial combustion.