At STP volume of an ideal gas is 22.4L/mol.

Therefore;

Volume of ethane mixed = `16L`

Amount of Ethane mixed = `16/22.4`

Mole ratio

`C_2H_6:CO_2 = 2:4 = 1:2`

Amount of `CO_2` produced = `(16/22.4)*2`

Volume of `CO_2` produced = `(16/22.4)*2*22.4 = 32 L`

**So there will be 32L of `CO_2` will be produced.**

*assumptions*

*`CO_2` and `C_2H_6` act as ideal gasses at the given conditions.*

**Further Reading**

Balanced chemical equation:

2C2H6 + 7O2 --> 4 CO2 + 6H2O

Using stoichiometry,

2 moles of ethane produce 4 moles of CO2.

Given volume of ethane gas = 16 lt.

At STP, 1 mole of a gas = 22.4 lt of gas

In the present case, 2 moles of ethane = 2x22.4 = 44.8 lts of ethane produces 4 x 22.4 lts (= 89.6 lts) of CO2

therefore, using unitary method,

16 l of ethane will produce = 89.6 x 16/44.8 lts of CO2 = **32 lts of CO2. **