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A

\(m_{HCl}=1,19.500.10=5950g\\ m_{ddHCl\left(37,23\%\right)}=\dfrac{5950}{37,23\%}=159,8g\\ V_{HCl}=159,8:1,19:1,19=112,86ml\)

vậy chọn A

.

\(m_{ddHCl\left(10\%\right)}=100.1,19=119g\\ m_{HCl}=119.10\%=1190g\\ m_{ddHCl\left(37,23\%\right)}=\dfrac{1190}{37,23}=32g\)

\(\Rightarrow V_{HCl}=32:1,19:1,19=22,6ml\)

\(a,n_{Zn}=\dfrac{19,5}{65}=0,3mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2}=n_{Zn}=0,3mol\\ V_{H_2}=0,3.24,79=7,437l\\ b,FeO+H_2\xrightarrow[t^0]{}Fe+H_2O\\ n_{FeO}=\dfrac{19,2}{72}=\dfrac{4}{15}mol\\ \Rightarrow\dfrac{0,3}{1}>\dfrac{4:15}{1}\Rightarrow H_2.dư\\ n_{Fe}=n_{FeO}=\dfrac{4}{15}mol\\ m_{Fe}=\dfrac{4}{15}.56\approx14,93g\)

Câu a tính V của cái gì á bạn?

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mik trl rồi bạn nhé!

a,

Số mol của Zn có trong pứ là:

\(\text{n}_{\text{Zn}}=\dfrac{\text{m}_{\text{Zn}}}{\text{M}_{\text{Zn}}}=\dfrac{19,5}{65}=0,3\left(\text{mol}\right)\)

PTHH: \(\text{Zn}+2\text{HCl}\longrightarrow\text{ZnCl}_2+\text{H}_2\)

Theo pt: 1  :  2  :  1  :  1

`=>` n của H₂ sau pứ là `0,3` mol

Thể tích của H₂ ở đkc là:

\(\text{V}_{\text{H}_2}=\text{n}_{\text{H}_2}\cdot24,79=0,3\cdot24,79=7,437\left(l\right)\)

b,

PTHH: \(\text{FeO}+\text{H}_2\) \(\underrightarrow{\text{ }\text{ t}^0\text{ }\text{ }}\) \(\text{Fe}+\text{H}_2\text{O}\)

n của FeO có trong pứ là:

\(n_{FeO}=\dfrac{m_{FeO}}{M_{FeO}}=\dfrac{19,2}{56+16}=\dfrac{19,2}{72}=\dfrac{4}{15}\left(\text{mol}\right)\)

Xét tỉ lệ:

\(\dfrac{0,3}{1}< \dfrac{4}{\dfrac{15}{1}}\Rightarrow\text{ H}_2\text{ dư}\)

Theo pt: 1   :   1

`=>` n của Fe sau pứ trên là `4/15` mol

Khối lượng của Fe sau pứ là:

\(\text{m}_{\text{Fe}}=\text{n}_{\text{Fe}}\cdot\text{M}_{\text{Fe}}=\dfrac{4}{15}\cdot56=\dfrac{224}{15}\approx14,93\left(\text{g}\right).\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{20\%}==73\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,2}{2,5}=0,08\left(l\right)\)

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

a, \(n_{H_2}=n_{Zn}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

Câu 1 : 

\(n_{Na2CO3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

Pt : \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O|\)

              1             1                 1            1           1

             0,1         0,1              0,1           0,1

a) \(n_{H2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{ddH2SO4}=\dfrac{0,1}{1}=0,1\left(l\right)\)

b) \(n_{CO2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{CO2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

c) \(n_{Na2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(C_{M_{Na2SO4}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

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