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

D.141 g

Step-by-step 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 = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (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