Question

If 49.0 g of O2 is mixed with 49.0 g of H2 and the mixture is ignited, what is the maximum mass of water that may be produced?

Answer 1

Answer:

$55.2gH_2O$

Explanation:

Hello!

In this case, since the reaction between hydrogen and oxygen to produce water is:

$2H_2+O_2\rightarrow 2H_2O$

When equal masses of each reactant go in contact to carry out the reaction, we can identify the maximum mass of yielded water as the fewest mass yielded by each reactant, just as shown below:

$m_{water}^{by\ H_2}=49.0gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molH_2O}{2molH_2}*\frac{18.02gH_2O}{1molH_2O} =170gH_2O\\\\m_{water}^{by\ O_2}=49.0gO_2*\frac{1molO_2}{32.0gO_2}*\frac{2molH_2O}{1molO_2}*\frac{18.02gH_2O}{1molH_2O} =55.2gH_2O$

Thus, since hydrogen yields more water than real, we limit the maximum mass of water to those 55.2 g yielded by oxygen as hydrogen would be in excess.

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