Question

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 23. g of ethane is mixed with 147. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

The minimum mass of ethane that could be left over by the chemical reaction is zero.

Explanation:

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

Moles of ethane = $\frac{23.0 g}{30 g/mol}=0.7666 mol$

Moles of oxygen gas :

$\frac{147.0 g}{32 g/mol}=4.59375 mol$

Since, 2 moles of ethane gas react with 7 moles of oxygen gas.

Then 0.7666 mol of ethane gas will react wit:

$\frac{7}{2}\times 0.7666 mol=2.6831 mol$

This means that ethane gas is in limited amount.hence ethane is a limiting reagent. Where as oxygen in present in excessive amount.

All the moles ethane gas will go under combustion reaction leaving zero moles behind.