Question

Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 7.9 g of methane is mixed with 14.4 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Explanation:

The given reaction is as follows.

     $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

As it is known that number of moles equal mass divided by molar mass. Molar mass of methane is 16 g/mol.

Hence,   No. of moles = $\frac{mass}{molar mass}$

                                   = $\frac{7.9 g}{16 g/mol}$

                                   = 0.493 mol

Molar mass of oxygen is 32 g/mol.

Hence, its number of moles = $\frac{mass}{molar mass}$

                                            = $\frac{14.4 g}{32 g/mol}$

                                            = 0.45 mol

As, it is shown from the reaction that 1 mole of methane needs 2 mole of oxygen.

Therefore, 0.493 mol of methane needs $2 \times 0.45 mol$ equals 0.9 moles of oxygen.

As there is only 0.45 moles of oxygen for the reaction. So, it means that oxygen is the limiting reagent.

Hence, for 0.45 moles of oxygen, methane required is as follows.

                        $\frac{0.45}{2}$ = 0.225 moles

So, 0.225 moles of methane is equal to $0.225 \times 16$ = 3.6 g

As, 3.6 g of methane reacts with oxygen. Therefore, amount of methane remains is calculated as follows.

                   (7.9 g - 3.6 g) = 4.3 g

 Thus, we can conclude that the minimum mass of methane that could be left over by the chemical reaction is 4.3 g.