Question

Outline the steps needed to determine the limiting reactant when 30.0 g of propane,\({C_3}{H_8}\), is burned with 75.0 g of oxygen. Determine the limiting reactant.

Short answer

Steps for limiting reactant.

1. Write the balanced chemical equation.
2. Find the molar mass of propane\({C_3}{H_8}\)and\({O_2}\).
3. Find the mass of \({O_2}\) required to burn 30.0g \({C_3}{H_8}\).
4. If the mass of \({O_2}\) present is more, \({O_2}\) is in excess. Hence, \({C_3}{H_8}\) is the limited reactant. If not, the \({O_2}\) is the limited reactant.

Step-by-step solution

Balanced chemical equation

Write the combustion of propane as

\(\)\(\)\({C_3}{H_8} + {O_2} \to C{O_2} + {H_2}O\)

Balance the equation by putting coefficients.

\({C_3}{H_8} + 5{O_2} \to 3C{O_2} + 4{H_2}O\)

Determine molar mass

\(\begin{aligned}{}1\,mol\,{C_3}{H_8} = 3\left( {12.011} \right) + 8\left( {1.008} \right)\\ = 44.097g\\1\,mol\,{O_2} = 2\left( {15.999} \right)\\ = 31.998g\end{aligned}\).

Determine mass of O2 required

Find the mass of \({O_2}\)required to burn 30.0g propane.

\(\begin{aligned}{}30.0\,g\,{C_3}{H_8}\left( {\frac{{1\,\,mol\,{C_3}{H_8}}}{{44.097\,g\,{C_3}{H_8}}}} \right)\left( {\frac{{5\,mol\,{O_2}}}{{1\,mol\,{C_3}{H_8}}}} \right)\left( {\frac{{31.998\,g\,{O_2}}}{{1\,mol\,{O_2}}}} \right) = 109\,g\,{O_2}\\75.0\,g{O_2} < 109\,g\,{O_2}\end{aligned}\)