a) ZnCl2+2KOH---->Zn(OH)2+2KCl
b)
m\(_{ZnCl2}=\frac{204.10}{100}=20,4\left(g\right)\)
n\(_{ }\)\(_{ZnCl2}=\frac{20,4}{136}=0,15\left(mol\right)\)
n\(_{KOH}=\frac{112.20\%}{56}=0,4\left(mol\right)\)
=> KOH dư
Theo pthh
n\(_{Cu\left(OH\right)2}=n_{ZnCl2}=0,15\left(mol\right)\)
m\(_{Cu\left(OH\right)2}=0,15.98=14,7\left(g\right)\)
c) m dd sau pư=204+112=316(g)
Theo pthh
n\(_{KOH}=2n_{ZnCl2}=0,3\left(mol\right)\)
C% KOH=\(\frac{0,3.56}{326}.100\%=5,32\%\)
n\(_{KCl}=2n_{ZnCl2}=0,3\left(mol\right)\)
C% KCl=\(\frac{0,3.74,5}{316}.100\%=7,07\%\)
\(a,ZnCL2+2KOH\rightarrow Zn\left(OH\right)2+2KCl\)
\(\text{b,nZnCl=204.10%/136=0,15(mol)}\)
\(\text{nKOH=112.20%/56=0,4(mol)}\)
=>KOH dư và nKOH dư=0,4-0,15.2=0,1(mol)
\(\text{nZn(OH)2=nZnCl2=0,15(mol)}\)
\(\Rightarrow mZn\left(OH\right)2=0,15.99=14,85\left(g\right)\)
\(\text{c) m dd spu=204+112=316(g)}\)\(\text{C%KOH dư= 0,1.56/316.100=1,772%}\)
\(\text{C%KCl=0,3.74,5/316.100=7,073%}\)
PTHH:ZnCl2+2KOH-->Zn(OH)2+2KCl
mol:1...........2......................1.............2
mol:0,15.....0,3..................0,15.........0,3
nZnCl2=\(\frac{201.10\%}{100\%.136}\)=0,15(mol)
nKOH=\(\frac{112.20\%}{100\%.56}\)=0,4(mol)
ta có tỉ lệ:
\(\frac{0,15}{1}\)<\(\frac{0,4}{2}\)=>KOH dư
mZn(OH)2=0,15.99=14,85(mol)
C%\(_{KCl}\)=\(\frac{0,3.74,45}{204+112}\).100%=1,07%
C%\(_{KOH}\)=\(\frac{\left(0,4-0,3\right).56}{112+204}\).100%=1,8%
C%\(_{KOHdu}\)nha bạn
a)ZnCl2+2KOH->Zn(OH)2+2KCl
b) nZnCl=204x10%/136=0,15(mol)
nKOH=112x20%/56=0,4(mol)
=>KOH dư và nKOH dư=0,4-0,15x2=0,1(mol) nZn(OH)2=nZnCl2=0,15(mol)
=>mZn(OH)2=0,15x99=14,85(g)
c) m dd spu=204+112=316(g)
C%KOH dư= 0,1x56/316x100=1,772% C%KCl=0,3x74,5/316x100=7,073%
