Question

A yield of 3.00kg \(KCL{O_4}\) is obtained from the unbalanced reaction\(KCl{O_3} \to KCl{O_4} + KCl\). When 4.00g of the reactant is used. What is the percentage yield of the reaction?

Short answer

Molar mass of \({\mathop{\rm K}\nolimits} = 39.10gmo{l^{ - 1}}\)

Molar mass of \({\mathop{\rm Cl}\nolimits} = 35.45gmo{l^{ - 1}}\)

Molar mass of \({\mathop{\rm O}\nolimits} = 16.00gmo{l^{ - 1}}\)

Molar mass of \({{\mathop{\rm KClO}\nolimits} _3}\) is \(\left( {39.10 + 35.45 + 3 \times 16.00} \right)gmo{l^{ - 1}} = 122.55gmo{l^{ - 1}}\)

Molar mass of \({{\mathop{\rm KClO}\nolimits} _4}\) is \(\left( {39.10 + 35.45 + 4 \times 16.00} \right)gmo{l^{ - 1}} = 138.55gmo{l^{ - 1}}\)

Step-by-step solution

Step 1:

Balance equation

• Unbalanced equation is

\({{\mathop{\rm KClO}\nolimits} _3} \to KCl{O_4} + KCl\)Assigning coefficient 4 to\({{\mathop{\rm KClO}\nolimits} _4}\). Therefore, coefficient of \({{\mathop{\rm KClO}\nolimits} _4}\)must be 3 to balance all the atoms of both sides

• Balanced equation is

\(4KCl{O_3} \to 3KCl{O_4} + KCl\)

Step 2:

Yield of\({{\mathop{\rm KClO}\nolimits} _4}\)

According to reaction 3 moles of\({{\mathop{\rm KClO}\nolimits} _4}\)is obtained from 4 moles of \({{\mathop{\rm KClO}\nolimits} _3}\)

Then amount of \({{\mathop{\rm KClO}\nolimits} _4}\) produced from 4.00g of \({{\mathop{\rm KClO}\nolimits} _3}\) is

\(\begin{aligned}{}4.00gKCl{O_3} \times \dfrac{{1molKCl{O_3}}}{{122.55gKCl{O_3}}} \times \dfrac{{3molKCl{O_3}}}{{4molKCl{O_3}}} \times \dfrac{{135.55gKCl{O_4}}}{{1molKCl{O_4}}}\\ = 3.39gKCl{O_4}\end{aligned}\)