Question

Đã trả lời bởi chuyên gia

Answer 1

MnO2+4HCl→MnCl2+Cl2+2H2OMnO2+4HCl→MnCl2+Cl2+2H2OMnO2+4HCl→MnCl2+Cl2+2H2O

Ta có: nMnO2=8,755+16.2=0,1 mol = nCl2nMnO2=8,755+16.2=0,1 mol = nCl2nMnO2=8,755+16.2=0,1 mol = nCl2

Zn+Cl2→ZnCl2Zn+Cl2→ZnCl2Zn+Cl2→ZnCl2

nZn=1365=0,2 mol > nCl2nZn=1365=0,2 mol > nCl2nZn=1365=0,2 mol > nCl2 nên Zn dư.

→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam

(mdd Z=100+13,6=113,6 gam→C%ZnCl2=13,6113,6=11,97%(mdd Z=100+13,6=113,6 gam→C%ZnCl2=13,6113,6=11,97%

Answer 2

Tham khảo nhé

Answer 3

MnO2+4HCl→MnCl2+Cl2+2H2O $M n O_{2} + 4 H C l \overset{}{\to} M n C l_{2} + C l_{2} + 2 H_{2} O$

Ta có: nMnO2=8,755+16.2=0,1 mol = nCl2 $n_{M n O_{2}} = \frac{8, 7}{55 + 16.2} = 0, 1 mol = n_{C l_{2}}$

Zn+Cl2→ZnCl2 $Z n + C l_{2} \overset{}{\to} Z n C l_{2}$

nZn=1365=0,2 mol > nCl2 $n_{Z n} = \frac{13}{65} = 0, 2 mol > n_{C l_{2}}$ nên Zn dư.

→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam $\to n_{Z n C l_{2}} = n_{C l_{2}} = 0, 1 mol \to m_{Z n C l_{2}} = 0, 1. (65 + 35, 5.2) = 13, 6 gam$

(mdd Z=100+13,6=113,6 gam→C%ZnCl2=13,6113,6=11,97% $(m_{d d Z} = 100 + 13, 6 = 113, 6 gam \to C %_{Z n C l_{2}} = \frac{13, 6}{113, 6} = 11, 97 %$

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Answer 4

MnO2+4HCl→MnCl2+Cl2+2H2O $M n O_{2} + 4 H C l \overset{}{\to} M n C l_{2} + C l_{2} + 2 H_{2} O$

Ta có: nMnO2=8,755+16.2=0,1 mol = nCl2 $n_{M n O_{2}} = \frac{8, 7}{55 + 16.2} = 0, 1 mol = n_{C l_{2}}$

Zn+Cl2→ZnCl2 $Z n + C l_{2} \overset{}{\to} Z n C l_{2}$

nZn=1365=0,2 mol > nCl2 $n_{Z n} = \frac{13}{65} = 0, 2 mol > n_{C l_{2}}$ nên Zn dư.

→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam $\to n_{Z n C l_{2}} = n_{C l_{2}} = 0, 1 mol \to m_{Z n C l_{2}} = 0, 1. (65 + 35, 5.2) = 13, 6 gam$

(mdd Z=100+13,6=113,6 gam→C%ZnCl2=13,6113,6=11,97% $(m_{d d Z} = 100 + 13, 6 = 113, 6 gam \to C %_{Z n C l_{2}} = \frac{13, 6}{113, 6} = 11, 97 %$

Answer 5

MnO2+4HCl→MnCl2+Cl2+2H2OMnO2+4HCl→MnCl2+Cl2+2H2OMnO2+4HCl→MnCl2+Cl2+2H2O

Ta có: nMnO2=8,755+16.2=0,1 mol = nCl2nMnO2=8,755+16.2=0,1 mol = nCl2nMnO2=8,755+16.2=0,1 mol = nCl2

Zn+Cl2→ZnCl2Zn+Cl2→ZnCl2Zn+Cl2→ZnCl2

nZn=1365=0,2 mol > nCl2nZn=1365=0,2 mol > nCl2nZn=1365=0,2 mol > nCl2 nên Zn dư.

→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam

(mdd Z=100+13,6=113,6 gam→C%ZnCl2=13,6113,6=11,97%(mdd Z=100+13,6=113,6 gam→C%ZnCl2=13,6113,6=11,97%

Answer 6

MnO2+4HCl→MnCl2+Cl2+2H2OMnO2+4HCl→MnCl2+Cl2+2H2OMnO2+4HCl→MnCl2+Cl2+2H2O

Ta có: nMnO2=8,755+16.2=0,1 mol = nCl2nMnO2=8,755+16.2=0,1 mol = nCl2nMnO2=8,755+16.2=0,1 mol = nCl2

Zn+Cl2→ZnCl2Zn+Cl2→ZnCl2Zn+Cl2→ZnCl2

nZn=1365=0,2 mol > nCl2nZn=1365=0,2 mol > nCl2nZn=1365=0,2 mol > nCl2 nên Zn dư.

→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam→nZnCl2=nCl2=0,1 mol→mZnCl2=0,1.(65+35,5.2)=13,6 gam

(mdd Z=100+13,6=113,6 gam→C%ZnCl2=13,6113,6=11,97%(mdd Z=100+13,6=113,6 gam→C%ZnCl2=13,6113,6=11,97%

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