carbon monoxide can be prepared by the reaction of steam with methane.
CH4(g) + H2O(g) = CO(g) + 3H2(g)
If 10g of methane is allowed to react with 10g of steam which of the reactants would be described as limiting. What is the theoretical yield of carbon monoxide if 10g of steam reacts?
I work limiting reagent problems by solving two simple stoichiometry problems (simple meaning not limiting reagent). Using one of the reactants solve for moles of the product formed. Then use the second reactant and solve for the same product. In limiting reagent problems, the correct answer is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is a solved example of solving a simple stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine which reactant is limiting, we need to compare the amounts of product that can be formed from each reactant.
First, let's calculate the number of moles of methane and steam using their respective molar masses:
Molar mass of CH4 (methane) = 12.01 g/mol (carbon) + 4 * 1.01 g/mol (hydrogen) = 16.05 g/mol
Molar mass of H2O (steam) = 2 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol
Number of moles of methane (CH4):
10 g CH4 * (1 mol CH4 / 16.05 g CH4) = 0.62 mol CH4
Number of moles of steam (H2O):
10 g H2O * (1 mol H2O / 18.02 g H2O) = 0.55 mol H2O
Now, let's use the balanced chemical equation to determine the mole ratio of CH4 to CO:
From the balanced equation: CH4(g) + H2O(g) = CO(g) + 3H2(g)
The mole ratio of CH4 to CO is 1:1.
Therefore, based on the above calculations, we can see that 0.62 mol of methane (CH4) is greater than the 0.55 mol of steam (H2O). This means that the steam is the limiting reactant.
To calculate the theoretical yield of carbon monoxide (CO), we need to use the mole ratio between the limiting reactant and the product.
From the balanced equation: CH4(g) + H2O(g) = CO(g) + 3H2(g)
The mole ratio of H2O to CO is 1:1.
So, 0.55 mol of H2O will theoretically produce 0.55 mol of CO.
To convert moles of CO to grams, we need to use the molar mass of CO:
Molar mass of CO = 12.01 g/mol (carbon) + 16.00 g/mol (oxygen) = 28.01 g/mol
Theoretical yield of CO:
0.55 mol CO * (28.01 g CO / 1 mol CO) = 15.4 g CO
Therefore, the theoretical yield of carbon monoxide (CO) when 10 g of steam reacts is 15.4 g CO.