Question

calculate the mass of ( p_{4}o_{10}(s) ) formed when 225 g of ( ph_{3} ) reacts with excess oxygen.

( 4ph_{3}(g)+8o_{2}(g)\to p_{4}o_{10}(s)+6h_{2}o(g) )

900 g
940 g
470 g
56.3 g
1880 g

Explanation:

Step1: Calculate the molar mass of \(PH_3\)

The molar mass of \(P = 30.97\ g/mol\) and \(H= 1.01\ g/mol\). So, \(M_{PH_3}=30.97+(3\times1.01)=34\ g/mol\)

Step2: Calculate the number of moles of \(PH_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 225\ g\) and \(M = 34\ g/mol\). So, \(n_{PH_3}=\frac{225}{34}\approx6.62\ mol\)

Step3: Use the stoichiometry of the reaction

From the balanced equation \(4PH_3(g)+8O_2(g)\to P_4O_{10}(s)+6H_2O(g)\), the mole ratio of \(PH_3\) to \(P_4O_{10}\) is \(4:1\). So, \(n_{P_4O_{10}}=\frac{n_{PH_3}}{4}=\frac{6.62}{4}\approx1.655\ mol\)

Step4: Calculate the molar mass of \(P_4O_{10}\)

The molar mass of \(P = 30.97\ g/mol\) and \(O = 16\ g/mol\). So, \(M_{P_4O_{10}}=(4\times30.97)+(10\times16)=283.88\ g/mol\)

Step5: Calculate the mass of \(P_4O_{10}\)

Using the formula \(m = n\times M\), where \(n = 1.655\ mol\) and \(M = 283.88\ g/mol\). So, \(m_{P_4O_{10}}=1.655\times283.88\approx470\ g\)

Answer:

470 g