$n_{Ba}=\frac{27,4}{137}=0,2 mol$

$n_{H_2SO_4}=\frac{200.14,7\%}{98}= 0,3 mol$

$Ba+H_2SO_4\to BaSO_4+H_2$

$\Rightarrow n_{BaSO_4}= n_{H_2}= n_{H_2SO_4 pứ}=0,2 mol$

$\Rightarrow n_{H_2SO_4 dư}=0,3-0,2=0,1 mol$

$m_{dd spứ}= m_{Ba}+ 200-m_{BaSO_4\downarrow} - m_{H_2\uparrow}$

$= 27,4+200-0,2.233-0,2.2=  180,4g$

$C\%_{H_2SO_4}=\frac{0,1.98.100}{180,4}= 5,4\%$

Đáp án:

 C% = 5,4%

Giải thích các bước giải:

\(n_{Ba}=\dfrac{27,4}{137}=0,2\ \text{mol}\\ n_{H_2SO_4}=\dfrac{200.14,7\%}{98}=0,3\ \text{mol}\\ Ba+H_2SO_4\to BaSO_4+H_2\)

`{0,2}/1<{0,3}/1->`Sau phản ứng Ba hết, \(H_2SO_4\) dư

\(\to n_{H_2SO_4\ \text{phản ứng}}=n_{Ba}=0,2\ \text{mol}\)

\(\to n_{H_2SO_4\ \text{dư}}=0,3-0,2=0,1\ \text{mol}\)

\(\to m_{H_2SO_4\ \text{dư}}=0,1.98=9,8\ \text{g}\)

\( n_{BaSO_4}=n_{Ba}=0,2\ \text{mol}\)

\(\to m_{BaSO_4}=233.0,2=46,6\ \text{gam}\)

\(n_{H_2}=n_{Ba}=0,2\ \text{mol}\to m_{H_2}=2.0,2=0,4\ \text{g}\)

\(\to m_{dd\ \text{sau phản ứng}}=m_{Ba}+200-m_{BaSO_4}-m_{H_2}=27,4+200-46,6- 0,4=180,4\ \text{g}\to C\%_{\text{dd sau phản ứng}}=\dfrac{9,8}{180,4}.100\%=5,4\%\)