Question

I need helping solving limiting reactant problems. I am confused on how to set up the equation(step 4).
How many grams of ZrCl4 can be produced if 123g ZrSiO4 react with 85g Cl2?
1.) ZrSiO4 + 2Cl2 -> ZrCl4 + SiO2 + O2
2.) 123g ZrSiO4 85g Cl2
3.) yes, limiting reactant problem
4.) This is where I get stuck. I know you set up two equations, and I know they both have to end with ZrCl4. One equations starts with 123g ZrSiO4 and the other starts with 85g Cl2 but I don't know what to do from there?

Answer 1

First find the moles of the each reactant. Divide each by it's stoichiometric coefficient in the balanced reaction e.g. \(\dfrac{moles~of~Cl_2}{2}\). From here, the ones with the least amount of moles is the limiting reactant. Proceed with the moles of limiting reactant (before you divided).

Use the stoichiometric coefficients to find moles of \(ZrCl_4\) produced.
Set up a ratio using the species of interest, like so:

e.g. for a general reaction:

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

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

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

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: \(\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)
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To convert mass to moles (or moles to mass), use the relationship:

\(n=\dfrac{m}{M}\)

where, M=molar mass, m=mass, and n= moles.