Question

If 0.40 mol of H2 and 0.15 mol of O2 were to react as completely as possible to produce H2O what mass of reactant would remain?

Answer 1

Answer : The mass of reactant $H_2$ remain would be, 0.20 grams.

Solution : Given,

Moles of $H_2$ = 0.40 mol

Moles of $O_2$ = 0.15 mol

Molar mass of $H_2$ = 2 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2H_2+O_2\rightarrow 2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $O_2$ react with 2 mole of $H_2$

So, 0.15 moles of $O_2$ react with $0.15\times 2=0.30$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

The moles of reactant $H_2$ remain = 0.40 - 0.30 = 0.10 mole

Now we have to calculate the mass of reactant $H_2$ remain.

$\text{ Mass of }H_2=\text{ Moles of }H_2\times \text{ Molar mass of }H_2$

$\text{ Mass of }H_2=(0.10moles)\times (2g/mole)=0.20g$

Therefore, the mass of reactant $H_2$ remain would be, 0.20 grams.

Answer 2

Answer:

The remaining mass of $\rm H_2$ in the reaction is 0.20 grams.

Explanation:

The balanced equation for the reaction will be:

$\rm 2\; H_2 \; + O_2 \rightarrow\; 2\; H_2O$

1 mole of $\rm O_2$ reacts with 2 moles of $\rm H_2$ to gives 2 moles of $\rm H_2O$.

0.15 moles of $\rm O_2$ reacts with 2 * 0.15 = 0.30 moles of $\rm H_2$

We have 0.40 moles of $\rm H_2$

So remaining $\rm H_2$= 0.40 moles - 0.30 moles

                         = 0.10 moles

Mass = moles * molar mass

Molar mass of $\rm H_2$ = 2 g/mol

Mass of $\rm H_2$ remained in the reaction = 0.10 moles * 2 g/mole

                                                            = 0.20 grams.

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