Cho 300ml dung dịch H2SO4, 5M và phản ứng với 400g dung dịch Ba(OH)2 8% a) Tính khối lượng kết tủa thu được b) Tính khối lượng muối tạo thành

2 câu trả lời

Đáp án:

$m_{BaSO_4}=43,571g$

Giải thích các bước giải:

 $n_{H_2SO_4}=0,3.5=1,5 mol$
$n_{Ba(OH)_2}=\frac{400.8}{100.171}= 0,187 mol$
phương trình phản ứng:

$H_2SO_4 + Ba(OH)_2 \to BaSO_4 + 2H_2O$
$n_{BaSO_4}=n_{Ba(OH)_2}=0,187 mol$
$m_{BaSO_4}=0,187.233=43,571g$

\(\begin{array}{l}
a)\\
300ml=0,3l\\
n_{H_2SO_4}=0,3.5=1,5(mol)\\
n_{Ba(OH)_2}=\frac{400.8\%}{171}\approx 0,187(mol)\\
H_2SO_4+Ba(OH)_2\to BaSO_4+2H_2O\\
Do\,n_{Ba(OH)_2}<n_{H_2SO_4}\to H_2SO_4\,dư\\
Theo\,PT:\,n_{BaSO_4}=n_{Ba(OH)_2}=0,187(mol)\\
\to m_{BaSO_4}=0,187.233=43,571(g)\\
b)\\
\rm Không\,có\,muối\,tạo\,thành\,spu
\end{array}\)

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