Question

If you start with 45.0 grams of ethylene and an excess of oxygen, how many grams of carbon dioxide will be produced?A.29 gB.57 gC.71 gD.141 g

Answer 1

Answer:

D.141 g

Explanation:

Given that:-

Mass of ethylene = 45.0 g

Molar mass of ethylene = 28.05 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{45.0\ g}{28.05\ g/mol}$

$Moles= 1.60\ mol$

According to the reaction below:-

$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

1 mole of ethylene produces 2 moles of carbon dioxide

So,

1.60 mole of ethylene produces 2*1.60 moles of carbon dioxide

Moles of carbon dioxide = 3.2 mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass = Moles*Molar mass = 3.2 mol x 44.01 g/mol = 141 g

D.141 g  of carbon dioxide will be produced