Question

g Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 0.96 g of methane is mixed with 6.37 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.

Answer 1

Answer: 2.64 g of carbon dioxide that could be produced by the chemical reaction.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} CH_4=\frac{0.96g}{16g/mol}=0.06moles$

$\text{Moles of} O_2=\frac{6.37}{32}=0.20moles$

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

According to stoichiometry :

1 mole of $CH_4$ require = 2 moles of $O_2$

Thus 0.06 moles of $CH_4$ will require=$\frac{2}{1}\times 0.06=0.12moles$  of $O_2$

Thus $CH_4$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

As 1 moles of $CH_4$ give = 1 mole of $CO_2$

Thus 0.06 moles of $CH_4$ give =$\frac{1}{1}\times 0.06=0.06moles$  of $CO_2$

Mass of $CO_2=moles\times {\text {Molar mass}}=0.06moles\times 44g/mol=2.64g$

Thus 2.64 g of carbon dioxide that could be produced by the chemical reaction.