Question

Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H₃PO₄ react according to the following chemical equation.\(2Cr + 2{H_3}P{O_4} \to 2CrP{O_4} + 3{H_2}\). Determine the limiting reactant.

Short answer

Cr is the limiting reactant.

Step-by-step solution

Determine the number of moles

Find the number of moles of \({H_3}P{O_4}\).

Mol\({H_3}P{O_4} = 0.50\,mol\,Cr\,\left( {\frac{{2\,mol\,{H_3}P{O_4}}}{{2\,mol\,Cr}}} \right) = 0.50\,mol\,{H_3}P{O_4}\).

Determine the limiting reactant

Only 0.50 mol of\({H_3}P{O_4}\)is required but we have 0.75 mol of\({H_3}P{O_4}\)so, \({H_3}P{O_4}\)is in excess amount. Thus, Cr is the limiting reactant.