# In the equation 2C_2H_2(5) + 5O_2(5) -> 4CO_2(g) + 2H_2O(g), how many grams of oxygen gas are needed for the complete combustion of 1.8 * 10^2 g of acetylene?

Jan 6, 2016

About $553.6$ grams of oxygen are used in this combustion.

#### Explanation:

Molecular mass of acetylene: $2 \cdot M {M}_{C} + 2 \cdot M {M}_{H}$
$M {M}_{C}$ = Molecular mass of carbon
$M {M}_{H}$ = Molecular mass of hydrogen

So, $2 \cdot 12 + 2 \cdot 1 = 26$ grams/mole

$\text{Mass used in combustion"/"molecular mass" = "moles used in combustion}$

(180cancel(" grams"))/(26 cancel(" grams")/"mole")= 6.92 " moles"

According to the main equation, to 2 moles of acetylene, 5 moles of oxygen are used. $M {M}_{O} = 2 \cdot 16 = 32$ grams/mole

So, if we use $6.92$ moles of Acetylene, $x$ moles of oxygen will be necessary.

$x = \frac{6.92 \cdot 5}{2} = 3.46 \cdot 5 = 17.3$ moles of oxygen.
$17.3 \cancel{\text{moles") * 32 "grams"/cancel("mole}} = 553.6$ grams of oxygen.