Question

For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be produced by the complete conversion of 0.50 mole of the reactant indicated in boldface. State clearly the mole ratio used for each conversion. a. NH3(g) 1 HCl(g) S NH4Cl(s)

Answer 1

Answer : The number of moles of $NH_4Cl$ and mass of $NH_4Cl$ is 0.50 mole and 27 grams respectively.

Solution : Given,

We are assuming that the reactant indicated in boldface is $NH_3$. So,

Moles of $NH_3$ = 0.50 mole

Molar mass of $NH_4Cl$ = 53.5 g/mole

First we have to calculate the moles of $NH_4Cl$

The balanced chemical reaction is,

$NH_3(g)+HCl(g)\rightarrow NH_4Cl(s)$

From the reaction, we conclude that

As, 1 mole of $NH_3$ react to give 1 mole of $NH_4Cl$

So, 0.50 moles of $NH_3$ react to give 0.50 moles of $NH_4Cl$

The number of moles of $NH_4Cl$  is, 0.50 moles.

Now we have to calculate the mass of $NH_4Cl$

$\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl$

$\text{ Mass of }NH_4Cl=(0.50moles)\times (53.5g/mole)=26.75g\approx 27g$

The mass of $NH_4Cl$  is, 27 grams.