Question

phosphorus can be prepared from calcium phosphate by the following reaction:
2ca₃(po₄)₂(s) + 6sio₂(s) + 10c(s) → 6casio₃(s) + p₄(s) + 10co(g)
phosphorite is a mineral that contains ca₃(po₄)₂ plus other non - phosphorus - containing compounds. what is the maximum amount of p₄ that can be produced from 1.1 kg of phosphorite if the phosphorite sample is 75% ca₃(po₄)₂ by mass? assume an excess of the other reactants.
mass =  g
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Explanation:

Step1: Calculate the mass of \(Ca_3(PO_4)_2\) in the sample

The mass of the phosphorite sample is \(m = 1.1\space kg=1100\space g\).
The mass percentage of \(Ca_3(PO_4)_2\) is \(75\%\).
So the mass of \(Ca_3(PO_4)_2\), \(m_{Ca_3(PO_4)_2}=1100\space g\times0.75 = 825\space g\)

Step2: Calculate the molar mass of \(Ca_3(PO_4)_2\)

The molar mass of \(Ca_3(PO_4)_2\):
\(M_{Ca_3(PO_4)_2}=3\times40 + 2\times(31 + 4\times16)=310\space g/mol\)

Step3: Calculate the number of moles of \(Ca_3(PO_4)_2\)

Using the formula \(n=\frac{m}{M}\), where \(m = 825\space g\) and \(M = 310\space g/mol\)
\(n_{Ca_3(PO_4)_2}=\frac{825\space g}{310\space g/mol}\approx2.66\space mol\)

Step4: Use the stoichiometry of the reaction

From the balanced equation \(2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\to6CaSiO_3(s)+P_4(s)+10CO(g)\)
The mole ratio of \(Ca_3(PO_4)_2\) to \(P_4\) is \(2:1\)
So the number of moles of \(P_4\), \(n_{P_4}=\frac{1}{2}n_{Ca_3(PO_4)_2}\)
\(n_{P_4}=\frac{1}{2}\times2.66\space mol = 1.33\space mol\)

Step5: Calculate the molar mass of \(P_4\)

The molar mass of \(P_4\), \(M_{P_4}=4\times31=124\space g/mol\)

Step6: Calculate the mass of \(P_4\)

Using the formula \(m = n\times M\), where \(n = 1.33\space mol\) and \(M = 124\space g/mol\)
\(m_{P_4}=1.33\space mol\times124\space g/mol\approx165\space g\)

Answer:

\(165\)