Question

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Answer 1

Bài 32:

\(n_{H_2SO_4}=\dfrac{180.19,6\%}{100\%}:98=0,36\left(mol\right)\\ n_{BaCl_2}=\dfrac{320.20,8\%}{100\%}:208=0,32\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\)

0,32<----0,32----->0,32-------->0,64

Xét \(\dfrac{0,36}{1}>\dfrac{0,32}{1}\Rightarrow n_{H_2SO_4.dư}=0,36-0,32=0,04\left(mol\right)\)

a. \(m_{BaSO_4}=0,32.233=74,56\left(g\right)\)

b. \(C\%_{HCl}=\dfrac{0,64.36,5.100\%}{180+320-74,56}=5,49\%\)

\(C\%_{H_2SO_4}=\dfrac{0,04.98.100\%}{180+320-74,56}=0,92\%\)

Bài 33:

\(n_{CuO}=\dfrac{1,6}{160}=0,01\left(mol\right)\\ n_{H_2SO_4}=\dfrac{30.9,8\%}{100\%}:98=0,03\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,01--->0,01----->0,01

Xét \(\dfrac{0,03}{1}>\dfrac{0,01}{1}\Rightarrow n_{H_2SO_4.dư}=0,03-0,01=0,02\left(mol\right)\)

\(C\%_{CuSO_4}=\dfrac{0,01.160.100\%}{1,6+30}=5,06\%\\ C\%_{H_2SO_4}=\dfrac{0,02.98.100\%}{1,6+30}=6,2\%\)

Answer 2

B34:

Gọi `x` là `n_(CuSO_4)` (x>0)

PTHH:

`CuSO_4+Ba(OH)_2\rightarrowBaSO_4+Cu(OH)_2`

    x------------------------->x----------->x

Có: \(m_{\downarrow}=m_{BaSO_4}+m_{Cu\left(OH\right)_2}=233x+98x=33,1\)

`\Rightarrowx=0,1`

Vậy số mol `CuSO_4` ban đầu là `0,1(mol)`

B35:

\(m_{AgCl}=m_{ddA}+m_{ddB}-m_{ddX}=170+150-305,65=14,35\left(g\right)\\ \Rightarrow n_{AgCl}=\dfrac{14,35}{143,5}=0,1\left(mol\right)\)

PTHH: 

\(AgNO_3+NaCl\rightarrow NaNO_3+AgCl\)

0,1<--------0,1<-----0,1<-------0,1

Trong A: \(C\%_{AgNO_3}=\dfrac{0,1.170.100\%}{170}=10\%\)

Trong B: \(C\%_{NaCl}=\dfrac{0,1.58,5.100\%}{150}=3,9\%\)

Trong X: \(C\%_{NaNO_3}=\dfrac{0,1.85.100\%}{305,65}=2,78\%\)

$HaNa$♬

Answer 3