When acetylene, C2H2, bums in oxygen, high temperatures are produced that are used for welding metals. 2C2H2(g) + 5O2(g) →^Δ 4CO2(g) + 2H2O(g) How many grams of CO2 are produced when 54.6 g of C2H2 is burned?

Question

When acetylene, [latex]C_{2}H_{2}[/latex], bums in oxygen, high temperatures are produced that are used for welding metals.

[latex] 2C_{2}H_{2}(g) + 5O_{2}(g) \xrightarrow[]{\Delta } 4CO_{2}(g) + 2H_{2}O(g)[/latex]

How many grams of [latex]CO_{2}[/latex] are produced when 54.6 g of [latex]C_{2}H_{2}[/latex] is burned?

Accepted Answer

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM	Given	Need	Connect
	54.6 g of [latex]C_{2}H_{2}[/latex]	grams of [latex]CO_{2}[/latex]	molar masses, mole- mole factor
	Equation
	[latex] 2C_{2}H_{2}(g) + 5O_{2}(g) \xrightarrow[]{\Delta } 4CO_{2}(g) + 2H_{2}O(g)[/latex]

STEP 2 Write a plan to convert the given to the needed quantity (moles).

[latex] grams of C_{2}H_{2} \boxed{\begin{array}{l} ext{Molar}\ ext{mass }\end{array}} → moles of C_{2}H_{2} \boxed{\begin{array}{l} ext{Mole-mole}\ ext{factor }\end{array}} → moles of CO_{2} \boxed{\begin{array}{l} ext{Molar}\ ext{mass }\end{array}} → grams of CO_{2}[/latex]

STEP 3 Use coefficients to write mole-mole factors.

[latex]\begin{array}{r c}\boxed{\begin{matrix} 1 mole of C_{2}H_{2}=26.04 g of C_{2}H_{2} \ \frac{26.04 g C_{2}H_{2} }{1 mole C_{2}H_{2}} ext{ and } \frac{1 mole C_{2}H_{2} }{26.04 g C_{2}H_{2}} \end{matrix}} & \boxed{\begin{matrix} 1 mole of CO_{2} = 44.01 g of CO_{2} \ \frac{44.01 g CO_{2} }{1 mole CO_{2} } ext{ and }\frac{1 mole CO_{2} }{44.01 g CO_{2}} \end{matrix}} \end{array} [/latex]

[latex] \boxed{\begin{array}{l} 2 moles of C_{2}H_{2} = 4 moles of CO_{2} \ \frac{2 moles C_{2}H_{2}}{4 moles CO_{2}} and \frac{4 moles CO_{2}}{2 moles C_{2}H_{2}} \end{array}} [/latex]

STEP 4 Set up the problem to give the needed quantity (moles).

[latex] \underset{Three SFs }{54.6 \cancel{g C_{2}H_{2}}} imes \overset{Exact}{\underset{Four SFs }{\boxed{\frac{1 \cancel{mole C_{2}H_{2}}}{26.04 \cancel{g C_{2}H_{2}}} }}} imes \overset{Exact}{\underset{Exact}{\boxed{\frac{4 \cancel{moles CO_{2}}}{2 \cancel{moles C_{2}H_{2}}} }}} imes \overset{Four SFs}{\underset{Exact}{\boxed{\frac{44.01 g CO_{2}}{1 \cancel{mole CO_{2}}} }}} = \underset{Three Sfs }{185 g of CO_{2}}[/latex]

Question 7.12: When acetylene, C2H2, bums in oxygen, high temperatures are ...

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