Question

Calculate the mass of butane needed to produce 72.9 grams of carbon dioxide

Answer 1

I got you 
The molecular formula for the combustion of butane in oxygen is:
2 C4H10 + 13 O2 ---> 8 CO2 + 10 H20

You take the mass of carbon dioxide, 56.8g, divide by its molar mass, 44.01g/mol, to produce the moles of carbon dioxide. This is multiplied by the molar ratio of butane/CO2, (2/8) = 1/4, which gives the moles of butane required to produce the carbon dioxide. Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g

Answer 2

Answer:

$24.02gC_{4}H_{10}$

Explanation:

The balanced equation is the following:

$_{2} C_{4}H_{10}+_{13}O_{2}=_{8}CO_{2}+_{10}H_{2}O$

The molecular mass of the carbon dioxide is:

Atomic mass of C = 12$\frac{g}{mol}$

Atomic mass of O = 16$\frac{g}{mol}$

Molecular mass of $CO_{2}$ = 12 + (16*2) = 44$\frac{g}{mol}$

The molecular mass of the butano is:

Atomic mass of C = 12$\frac{g}{mol}$

Atomic mass of H = 1$\frac{g}{mol}$

Molecular mass of $C_{4}H_{10}$ = (4*12) + (10*1) = 58$\frac{g}{mol}$

Find the grams of carbon dioxide using the stoichiometry of the reaction:

$72.9gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}}*\frac{2molesC_{4}H_{10}}{8molesCO_{2}}*\frac{58gC_{4}H_{10}}{1molC_{4}H_{10}}=24.02gC_{4}H_{10}$