Correct Let's calculate the number of moles of each reactant: 1 mol CH4 1370 g CH4X = 85.4 mol CH, 16.04 g CH4 1 mol H20 18.02 g H20 1280 g H,0 x = 71.0 mol H20 Now compare the number of moles of the reactants, taking into account the fact that each mole of CH4 requires 1 mole of H20 to react completely: 71.0 mol H2 O 0.832 mol H20 85.4 mol CH, 1 mol CH4 1 mol H20 1 mol CH4 0.832 mol H2 O 1 mol CH4 So, H20 is the limiting reactant. b what is the maximum mass of H2 that can be prepared? Mass = g H2 Submit

Correct Let's calculate the number of moles of each reactant: 1 mol CH4 1370 g CH4X = 85.4 mol CH, 16.04 g CH4 1 mol H20 18.02 g H20 1280 g H,0 x = 71.0 mol H20 Now compare the number of moles of the reactants, taking into account the fact that each mole of CH4 requires 1 mole of H20 to react completely: 71.0 mol H2 O 0.832 mol H20 85.4 mol CH, 1 mol CH4 1 mol H20 1 mol CH4 0.832 mol H2 O 1 mol CH4 So, H20 is the limiting reactant. b what is the maximum mass of H2 that can be prepared? Mass = g H2 Submit

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter3: Molecules, Moles, And Chemical Equations

Section: Chapter Questions

Problem 3.79PAE: 3.79 Consider two samples. Sample A contains 2 moles of N2 and 1 mole of O2, and Sample B contains 1...

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