# Question #8198e

Dec 15, 2017

7.48 g

#### Explanation:

Convert volume of methane in the balloon to moles:

Density methane = 0.716 g/L at STP
3.75 x 0.716 = 2.69 g of methane

Molar Mass $C {H}_{4}$ = 16.04 g/mol

Using the equation: mol = mass / molar mass

Number of moles $C {H}_{4}$ = 2.69 / 16.04
Number of moles $C {H}_{4}$ = 0.17 mol

We need to write a balanced equation for this reaction to determine the molar ratio of $C {O}_{2}$ and $C {H}_{4}$

$C {H}_{4} + 2 {O}_{2} \to C {O}_{2} + 2 {H}_{2} O$

$C {O}_{2}$ and $C {H}_{4}$ in 1 : 1 ratio, so 0.17 moles of $C {O}_{2}$ are produced in the combustion reaction.

Number of moles of $C {O}_{2}$ = 0.17 mol
Molar mass $C {O}_{2}$ = 44.01 g/mol

Rearranging the equation we used before: mass = mol x molar mass
Mass $C {O}_{2}$ = 44.01 x 0.17
Mass $C {O}_{2}$ = 7.48 g