Question

The mass of ammonia in grams produced when $2.8\mathrm{kg}$of dinitrogen quantitatively reacts with $1\mathrm{kg}$ of dihydrogen is

Answer 1

Step 1 : Given data

$2.8\mathrm{kg}$ of dinitrogen quantitatively reacts with $1\mathrm{kg}$ of dihydrogen

${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$

$2800\mathrm{g}$ of dinitrogen reacts with $1000\mathrm{g}$ of dihydrogen

Step 2 : To find the number of moles in dinitrogen and dihydrogen:

Number of moles of dinitrogen $=\frac{2800}{28}$

Then the number of moles of dinitrogen =100

Number of moles of dihydrogen $=\frac{1000}{2}$

Number of moles of dihydrogen$=500$

$1$ mole of $N_2$ requires $3$ moles of $H_2$

$N_2$is a limiting reagent because dihydrogen is in excess as only $300$ moles of dihydrogen can react with $100$ moles of dinitrogen and $200$ moles of hydrogen in excess

Step 3: Product formation reaction:

According to the reaction

$1$ mole of $N_2$ gives $2$ moles of ammonia

$100$ mole of $N_2$gives $200$moles of ammonia

The molecular mass or mass of one mole of ammonia is $17$ grams

Mass of $200$ moles of ammonia $=200\times 17$

Mass of $200$ moles of ammonia $=3400grams$

Hence the mass of ammonia in grams produced when $2.8kg$ of dinitrogen quantitively reacts with $1kg$for $200\times 17$ is $3400grams$