Question

given the following reaction:
p₄o₁₀ + 6 h₂o → 4 h₃po₄
if 80.41 g of p₄o₁₀ is reacted with an excess of water, what is the maximum mass of phosphoric acid (h₃po₄) that could be produced?
111.0 g
20.22 g
7.584 g
53.29 g

Explanation:

Step1: Calculate the molar mass of \(P_4O_{10}\)

The molar mass of \(P = 30.97\space g/mol\), \(O=16.00\space g/mol\).
For \(P_4O_{10}\): \(M=(4\times30.97 + 10\times16.00)\space g/mol=(123.88+160)\space g/mol = 283.88\space g/mol\)

Step2: Calculate the number of moles of \(P_4O_{10}\)

Using the formula \(n=\frac{m}{M}\), where \(m = 80.41\space g\) and \(M = 283.88\space g/mol\)
\(n=\frac{80.41\space g}{283.88\space g/mol}\approx0.2833\space mol\)

Step3: Use the stoichiometry of the reaction

From the reaction \(P_4O_{10}+6H_2O
ightarrow4H_3PO_4\), the mole ratio of \(P_4O_{10}\) to \(H_3PO_4\) is \(1:4\).
So the number of moles of \(H_3PO_4\) produced \(n_{H_3PO_4}=4\times n_{P_4O_{10}}\)
\(n_{H_3PO_4}=4\times0.2833\space mol = 1.1332\space mol\)

Step4: Calculate the molar mass of \(H_3PO_4\)

The molar mass of \(H = 1.008\space g/mol\), \(P = 30.97\space g/mol\), \(O = 16.00\space g/mol\)
\(M_{H_3PO_4}=(3\times1.008+30.97 + 4\times16.00)\space g/mol=(3.024 + 30.97+64)\space g/mol=97.994\space g/mol\)

Step5: Calculate the mass of \(H_3PO_4\)

Using the formula \(m = n\times M\), where \(n = 1.1332\space mol\) and \(M = 97.994\space g/mol\)
\(m=1.1332\space mol\times97.994\space g/mol\approx111.0\space g\)

Answer:

111.0 g