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?

Question

Carbon monoxide and hydrogen are used to produce methanol [latex](CH_{4}O)[/latex]. The balanced chemical reaction is

[latex] CO(g) + 2H_{2}(g) → CH_{4}O(g) [/latex]

If 3.00 moles of CO and 5.00 moles of [latex]H_{2}[/latex] are the initial reactants, what is the limiting reactant, and how many moles of methanol can be produced?

Accepted Answer

STEP 1 State the given and needed quantities (moles).

ANALYZE THE PROBLEM	Given	Need	Connect
	3.00 moles of CO, 5.00 moles of [latex]H_{2}[/latex]	moles of [latex]CH_{4}O[/latex] produced, limiting reactant	mole-mole factors
	Equation
	[latex] CO(g) + 2H_{2}(g) → CH_{4}O(g) [/latex]

STEP 2 Write a plan to convert the quantity (moles) of each reactant to quantity (moles) of product.

[latex] moles of CO \boxed{\begin{array}{l} ext{Mole-mole}\ ext{factor}\end{array}} → moles of CH_{4}O[/latex]

[latex] moles of H_{2} \boxed{\begin{array}{l} ext{Mole-mole}\ ext{factor}\end{array}} → moles of CH_{4}O[/latex]

STEP 3 Use coefficients to write mole-mole factors.

[latex] \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} ext{ 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} ext{ and }\frac{ 1 mole CH_{4}O}{2 moles H_{2}} \end{matrix}}\end{array} [/latex]

STEP 4 Calculate the quantity (moles) of product from each reactant, and select the smaller quantity (moles) as the limiting reactant.
Moles of [latex]CH_{4}O[/latex] (product) from CO:

[latex] \underset{Three SFs }{3.00 \cancel{moles CO}} imes \overset{Exact}{\underset{Exact}{\boxed{\frac{1 mole CH_{4}O}{1 \cancel{ mole CO}} }}} = \underset{Three Sfs }{3.00 moles of CH_{4}O}[/latex]

Moles of [latex]CH_{4}O[/latex] (product) from [latex]H_{2}[/latex] :

[latex] \overset{Three SFs}{\underset{Limiting reactant }{5.00 \cancel{moles H_{2}}}} imes \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} ext{Smaller amount}\ ext{of product}\end{array} [/latex]

The smaller amount, 2.50 moles of [latex]CH_{4}O[/latex], is the maximum amount of methanol that can be produced from the limiting reactant, [latex]H_{2}[/latex] , because it is completely consumed.

Question 7.13: Carbon monoxide and hydrogen are used to produce methanol (C...

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