Question

What mass of carbon would be required to produce .20 mol of P4 in the presence of 1.0 mol of calcium phosphate and 3.0 mole of silicon dioxide?

2Ca3(PO4)2 + 10C + 6SiO2 -> P4 + 6CaSiO3 + 10 CO

Answer 1

Typically, you would need to find the "limiting reactant" - the reactant that has the least moles taking into account the stoichiometric ratio (the numbers infront of each reactant in the balanced equation). But here, since you're already given the moles of product \((P_4)\) you can skip that step.

Set up a ratio using the species of interest, like so:

e.g. for a general reaction:

\(\sf \color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients ,

\(\sf \dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

From here you can isolate what you need.

For example: if you have 2 moles of B, how many moles of C can you produce?

solve algebraically: \(\sf\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)



Your equation is this: \(\sf 2~Ca_3(PO_4)_2 + 10~C + 6~SiO_2 -> P4 + 6~CaSiO_3 + 10 ~CO\)

So the ratio you're using is between moles of C and P.

\(\sf \dfrac{n_C}{10}=\dfrac{n_{P_4}}{1}\)

Solve for moles of P and convert them to mass with the molar mass.

\(\sf moles =\dfrac{mass}{Molar ~mass}\)