Question

Gaseous methane CH4 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 0.963 g of methane is mixed with 1.2 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

Answer:

$m_{H_2O}=0.676gH_2O$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the maximum mass of water by firstly setting up the undergoing chemical reaction as follows:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Thus, we are able to firstly calculate the moles of water produced by both methane and oxygen in order to identify the limiting reactant, which is related to maximum of water:

$0.963 gCH_4*\frac{1molCH_4}{16gCH_4}*\frac{2molH_2O}{1molCH_4} =0.120molH_2O\\\\1.2gO_2*\frac{1molO_2}{32gO_2}*\frac{2molH_2O}{2molO_2} =0.0375molH_2O$

Thus, we infer the limiting reactant is O2 and therefore we can obtain up to 0.0375 moles of water, which are related to the following mass:

$m_{H_2O}=0.0375molH_2O\frac{18.02gH_2O}{1molH_2O}=0.676gH_2O$

Regards!