The amount of \(\mathrm{BaSO}_{4}\) formed upon mixing \(100 \mathrm{ml}\) of \(20.8 \% \mathrm{BaCl}_{2}\)…
The amount of \(\mathrm{BaSO}_{4}\) formed upon mixing \(100 \mathrm{ml}\) of \(20.8 \% \mathrm{BaCl}_{2}\) solution with 50 \(\mathrm{ml}\) of \(9.8 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) solution will be:
(Given that molecular weight of \(\mathrm{Ba}=137, \mathrm{Cl}=35.5, \mathrm{~S}=32, \mathrm{H}=1\) and \(\mathrm{O}=16\) \(\mathrm{g} / \mathrm{mol})\)
\(23.3 \mathrm{~g}\)
\(11.65 \mathrm{~g}\)
\(30.6 \mathrm{~g}\)
\(33.2 \mathrm{~g}\)
Solution
\(100 \mathrm{ml}\) of \(20.8 \% \mathrm{BaCl}_{2}\) solution \(=20.8 \mathrm{~g} \mathrm{BaCl}_{2}=0.1 \mathrm{~mol}\).
\(50 \mathrm{ml}\) of \(9.8 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) solution \(=4.9 \mathrm{~g} \mathrm{H}_{2} \mathrm{SO}_{4}=0.05 \mathrm{~mol}\).
The reaction is as follows:
\(\mathrm{BaCl}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \leftrightharpoons \mathrm{BaSO}_{4}+2 \mathrm{HCl}\)
Since, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is the limiting reagent, only \(0.05\) moles of \(\mathrm{BaSO}_{4}\) will form.
Hence, \(0.05 \times 233=11.65 \mathrm{~g}\) of \(\mathrm{BaSO}_{4}\) is formed.
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