Q68E - Chemistry (OpenStax)

Question

In a laboratory experiment, the reaction of 3.0 mol of \({H_2}\) with 2.0 mol of \({I_2}\) produced 1.0 mol of HI. Determine the theoretical yield in grams and the percent yield for this reaction.

Step-by-Step Solution

Answer

The percent yield is\(25\% \).

The theoretical yield is 511.6g HI.

Determine theoretical yield

1 mol HI\( = 1.008 + 126.904 = 127.912g\)

\(2.0{\rm{ }}mol{\rm{ }}{I_2}\left( {\frac{{2molHI}}{{1mol{I_2}}}} \right)\left( {\frac{{127.912gHI}}{{1molHI}}} \right) = 511.6gHI\)

\(1.0{\rm{ }}mol{\rm{ HI}}\left( {\frac{{127.912g\,HI}}{{1mol\,HI}}} \right) = 127.9g\,HI\)

Determine the percent yield

Calculate the percent yield.

%yield = \(\frac{{{\rm{actual yield}}}}{{{\rm{theoretical yield}}}} \times 100\% \)

             \(\begin{aligned}{\underline{\phantom{xx}}} &= \frac{{127.9g}}{{511.6g}} \times 100\\ &= 25\% \end{aligned}\)