Question

Plz help will give brainliest, 50 pts!!! use your knowledge of the law of conversation of mass to answer the following question: From a laboratory process, a student collects 28.0g of hydrogen and 224.0g of oxygen. How much water was originally involved in the process?

Answer 1

According to law of conservation of mass, in a chemical reaction mass can neither be destroyed nor created thus, the total mass created here will be 224 g+28 g=252 g thus, mass of $H_{2}O$ involved should be 252 g.

This can be experimentally proved as follows:

The balanced chemical reaction for the above process will be as follows:

$2H_{2}O\rightarrow 2H_{2}+O_{2}$

Considering any one of the reactant.

2 mol of $H_{2}$ are produced from 2 mol of $H_{2}O$ thus, 1 mol of $H_{2}$ produced from 1 mol of $H_{2}O$.

Now, mass of hydrogen gas is 28.0 g, molar mass is 2 g/mol, converting mass into number of moles:

$n=\frac{m}{M}=\frac{28.0 g}{2 g/mol}=14 mol$

thus, 14 mol of $H_{2}$ produced from 1×14=14 mol of $H_{2}O$.

Molar mass of $H_{2}O$ is 18 g/mol. Converting number of moles into mass as follows:

m=n×M=14 mol×18 g/mol=252 g

Thus, mass of $H_{2}O$ involved in the reaction will be 252 g.