Question 7.13: Carbon monoxide and hydrogen are used to produce methanol (C...
Carbon monoxide and hydrogen are used to produce methanol (CH_{4}O). The balanced chemical reaction is
CO(g) + 2H_{2}(g) → CH_{4}O(g)If 3.00 moles of CO and 5.00 moles of H_{2} are the initial reactants, what is the limiting reactant, and how many moles of methanol can be produced?
STEP 1 State the given and needed quantities (moles).
| ANALYZE THE PROBLEM | Given | Need | Connect |
| 3.00 moles of CO, 5.00 moles of H_{2} | moles of CH_{4}O produced, limiting reactant | mole-mole factors | |
| Equation | |||
| CO(g) + 2H_{2}(g) → CH_{4}O(g) | |||
STEP 2 Write a plan to convert the quantity (moles) of each reactant to quantity (moles) of product.
moles of CO \boxed{\begin{array}{l}\text{Mole-mole}\\\text{factor}\end{array}} → moles of CH_{4}O moles of H_{2} \boxed{\begin{array}{l}\text{Mole-mole}\\\text{factor}\end{array}} → moles of CH_{4}OSTEP 3 Use coefficients to write mole-mole factors.
\begin{array}{r c}\boxed{\begin{matrix} 1 mole of CO = 1 mole of CH_{4}O \\\frac{1 mole CO}{1 mole CH_{4}O} \text{ and } \frac{1 mole CH_{4}O}{1 mole CO} \end{matrix}} & \boxed{\begin{matrix} 2 moles of H_{2} = 1 mole of CH_{4}O\\ \frac{2 moles H_{2}}{ 1 mole CH_{4}O}\text{ and }\frac{ 1 mole CH_{4}O}{2 moles H_{2}} \end{matrix}}\end{array}STEP 4 Calculate the quantity (moles) of product from each reactant, and select the smaller quantity (moles) as the limiting reactant.
Moles of CH_{4}O (product) from CO:
Moles of CH_{4}O (product) from H_{2} :
\overset{Three SFs}{\underset{Limiting reactant }{5.00 \cancel{moles H_{2}}}} \times \overset{Exact}{\underset{Exact}{\boxed{\frac{1 mole CH_{4}O}{2 \cancel{ moles H_{2}}} }}} = \underset{Three Sfs }{2.50 moles of CH_{4}O} \begin{array}{l} \text{Smaller amount}\\ \text{of product}\end{array}The smaller amount, 2.50 moles of CH_{4}O, is the maximum amount of methanol that can be produced from the limiting reactant, H_{2} , because it is completely consumed.