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![](https://rs.olm.vn/images/avt/0.png?1311)
Ptpư Zn + 2HCl ---> ZnCl2 + H2
nH2 = 1 mol
mH2 = 2(g)
Áp dụg ĐLBTKL
mHCl = mZnCl2 +mH2 - mZn = 136 +2-65= 73(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + H2SO4 ---> ZnSO4 + H2
0,2--->0,2--------->0,2------>0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{H_2SO_4}=\dfrac{0,2.98}{20\%}=98\left(g\right)\\ \rightarrow V_{ddH_2SO_4}=\dfrac{98}{1,14}=86\left(ml\right)=0,086\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,086}=2,33M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mddH2SO4=4,9%.200=9,8(g)
-> nH2SO4=9,8/98=0,1(mol)
PTHH: Zn + H2SO4 ->ZnSO4 + H2
nH2=nH2SO4=0,1(mol)
=> V(H2,đktc)=0,1.22,4=2,24(l)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo ĐLBTKL:
mZn + mHCl = mH2 + mZnCl2
=> mHCl = 40,8 + 0,3.2 - 19,5 = 21,9(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 < 0,4 ( mol )
0,1 0,2 0,1 ( mol )
Chất dư là HCl
\(n_{HCl\left(dư\right)}=0,4-0,2=0,2mol\)
\(V_{H_2}=0,1.22,4=2,24l\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{NaOH}=\dfrac{6}{40}=0,15\left(mol\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot0,15=0,075\left(mol\right)\\ \Rightarrow m=m_{H_2SO_4}=0,075\cdot98=7,35\left(g\right)\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ \Rightarrow n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V=V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
Theo ĐLBTKL: mZn + mH2SO4 = mZnSO4 + mH2
=> mH2 = 3,25 + 4,9 - 8,05 = 0,1(g)