g Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 2.71 g of ethane is mixed with 3.8 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

6.05g

Explanation:

The reaction is given as;

Ethane + oxygen - -> Carbon dioxide + water

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

From the reaction above;

2 mol of ethane reacts with 7 mol of oxygen.

To proceed, we have to obtain the limiting reagent,

2,71g of ethane;

Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol

3.8g of oxygen;

Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol

If 0.0903 moles of ethane was used, it would require;

2 = 7

0.0903 = x

x = 0.31605 mol of oxygen needed

This means that oxygen is our limiting reagent.

From the reaction,

7 mol of oxygen yields 4 mol of carbon dioxide

0.2375 yields x?

7 = 4

0.2375 = x

x = 0.1357

Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g