How many molecules of C2H4Cl2 can be prepared from 15 C2H4 molecules and 8 Cl2 molecules?

8 molecules of \({C_2}{H_4}C{l_2}\) can be prepared from 15 C2H4 molecules and 8 Cl2 molecules

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How many molecules of C₂H₄Cl₂ can be prepared from 15 C₂H₄molecules and 8 Cl₂molecules?

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Answer

8 molecules of \({C_2}{H_4}C{l_2}\) can be prepared from 15 C₂H₄molecules and 8 Cl₂molecules

1Given data

Molecule \(C{l_2}\)must react with 1 molecule of \({C_2}{H_4}\), so 8 \(C{l_2}\) molecule can react with 8 molecule of \({C_2}{H_4}\).

2Determine theoretical yield

Find the theoretical yield.

\(8\,mol\,C{l_2}\left( {\frac{{1mol\,{C_2}{H_4}C{l_2}}}{{1molC{l_2}}}} \right) = 8mol\,{C_2}{H_4}C{l_2}\).