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

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Let's calculate the number of moles of each reactant:
1 mol CH4
1370 g CH, x
= 85.4 mol CH4
16.04 g CH4
1 mol H2 O
18.02 g H20
1280 g H,O x
71.0 mol H20
%3D
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 H20
0.832 mol H2 O
85.4 mol CH4
1 mol CH4
1 mol H20
1 mol CH4
0.832 mol H2О
1 mol CH,
So, H20 is the limiting reactant.
b what is the maximum mass of H, that can be prepared?
Mass =
g H2
Submit
Transcribed Image Text:Correct Let's calculate the number of moles of each reactant: 1 mol CH4 1370 g CH, x = 85.4 mol CH4 16.04 g CH4 1 mol H2 O 18.02 g H20 1280 g H,O x 71.0 mol H20 %3D 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 H20 0.832 mol H2 O 85.4 mol CH4 1 mol CH4 1 mol H20 1 mol CH4 0.832 mol H2О 1 mol CH, So, H20 is the limiting reactant. b what is the maximum mass of H, that can be prepared? Mass = g H2 Submit
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