Question

Q.7 The amount of BaSo, formed upon mixing 100 mL of 20.8% BaCl, solution with 50 mL of 9.8% H,SO
solution will be :
(JEE(Main-online)-2014]
(Ba = 137, CI = 35.5, S=32, H = 1 and 0 = 16)​

Answer 1

$\bf \underline \blue{Question : - }$

The amount of BaSo, formed upon mixing 100 mL of 20.8% BaCl, solution with 50 mL of 9.8% H,SO

solution will be ?

$\bf \underline \blue{Answer : - }$

$\star\bf \: 100 \: \: ml \: \: of \: \: 20.8 \: \: \: BaCl_2 \: \: Solution \: \: = 20.8 \: \: g \: \: BaCl_2 \\ \\ \bf \implies \: 0.1 \: \: mol$

$\star \bf \: 50 \: \: ml \: \: of \: \: 9.8 \: \: H_2SO_4 \: \: Solution \: \: = \: \: 4.9 \: \: g \: \: \: H_2SO_4 \\ \\ \bf \implies \: 0.05 \: \: mol$

$\sf \large \underline{ \: the \: \: reaction \: \: was \: \: given \: \: below \: \: : - }$

$\boxed{ \red { \bf \: BaCl_2+H_2SO_4⇋BaSO_4+2HCl}}$

$\sf \: Since \: \: H_2SO_4 \: \: is \: \: the \: \: limiting \: \: reagents, \: \: only \: \: 0.05 \: \: \: moles \: \: of \: \: BaSO_4 \: \: will \: \: form,$

$\tt \purple{0.05 \: × \: 233 \: = \: 11.65 \: \: g \: \: of \: \: BaSO_4 \: \: formed!!!!}$