Đáp án:
a) Viết ptpứ xảy ra: `H_2SO_4+BaCl_2toBaSO_4↓+2Hcl`
`mH_2SO_4={20\times29,4}/100=5,88(g)`
Suy ra `nH_2SO_4=0,06(mol)`
`mBaCl_2={5,2\times100}/100=5,2(g)`
Suy ra 'nBaCl_2=0,025(mol)`
$\dfrac{n_{H_{2}SO_4} \text{pứ}}{n_{H_{2}SO_4} \text{bđ}}=$`{0,06}/1>`$\dfrac{n_{BaCl_2} \text{pứ}}{n_{BaCl_2}\text{bđ}}=$`{0,025}/1`
Suy ra `H_2SO_4` dư, `BaCl_2` hết
Số `nBaCl_2=nBaSO_4(=0,025mol)`
Suy ra `mBaSO_4=0,025\times233=5,825(g)`
b) Những chất còn lại sau pứ là `H_2SO_4` dư, `Hcl`.
`mdd` sau pứ `=29,4+100-5,825=123,575(g)`
Số `nH_2SO_4` dư `=0,06-0,025=0,035(mol)`
`mH_2SO_4=0,035\times98=3,43(g)`
Suy ra `C%H_2SO_4={mct}/{mdd}\times100={3,43}/{123,575}\times100≈2,78%`
Số `nHcl=0,025\times2=0,05(mol)`
`mHcl=0,05\times36,5=1,825(g)`
Suy ra `C%Hcl={mct}/{mdd}\times100={1,825}/{123,575}\times100≈1,47%`
