Question

The reaction of NH3 and O2 forms NO and water. The NO can be used to convert P4 to P4O6, forming N2 in the process. The P4O6 can be treated with water to form H3PO3, which forms PH3 and H3PO4 when heated. Find the mass of PH3 that forms from the reaction of 1.95 g NH3.

Answer 1

Answer : The mass of $PH_3$ produced from the reaction is, 0.651 grams.

Explanation :

The chemical reactions used are:

(1) $4NH_3+5O_2\rightarrow 4NO+6H_2O$

(2) $6NO+P_4\rightarrow P_4O_6+3N_2$

(3) $P_4O_6+6H_2O\rightarrow 4H_3PO_4$

(4) $4H_3PO_4\rightarrow PH_3+3H_3PO_4$

First we have to calculate the moles of $NH_3$

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}$

Molar mass of $NH_3$ = 17 g/mol

$\text{Moles of }NH_3=\frac{1.95g}{17g/mol}=0.115mol$

Now we have to calculate the moles of $NO$

From the balanced chemical reaction 1, we conclude that:

As, 4 moles of $NH_3$ react to give 4 moles of $NO$

So, 0.115 moles of $NH_3$ react to give 0.115 moles of $NO$

Now we have to calculate the moles of $P_4O_6$

From the balanced chemical reaction 2, we conclude that:

As, 6 moles of $NO$ react to give 1 moles of $P_4O_6$

So, 0.115 moles of $NO$ react to give $\frac{0.115}{6}=0.0192$ moles of $P_4O_6$

Now we have to calculate the moles of $H_3PO_4$

From the balanced chemical reaction 3, we conclude that:

As, 1 moles of $P_4O_6$ react to give 4 moles of $H_3PO_4$

So, 0.0192 moles of $P_4O_6$ react to give $0.0192\times 4=0.0768$ moles of $H_3PO_4$

Now we have to calculate the moles of $PH_3$

From the balanced chemical reaction 4, we conclude that:

As, 4 moles of $H_3PO_4$ react to give 1 moles of $PH_3$

So, 0.0768 moles of $H_3PO_4$ react to give $\frac{0.0768}{4}=0.0192$ moles of $PH_3$

Now we have to calculate the mass of $PH_3$

$\text{ Mass of }PH_3=\text{ Moles of }PH_3\times \text{ Molar mass of }PH_3$

Molar mass of $PH_3$ = 33.9 g/mole

$\text{ Mass of }PH_3=(0.0192moles)\times (33.9g/mole)=0.651g$

Therefore, the mass of $PH_3$ produced from the reaction is, 0.651 grams.