Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(m_{CuSO_4}=40.10\%=4\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{4}{160}=0,025\left(mol\right)\)
PT: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)
Theo PT: \(n_{Zn}=n_{ZnSO_4}=n_{Cu}=n_{CuSO_4}=0,025\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,025.65=1,625\left(g\right)\)
Ta có: m dd sau pư = 1,625 + 40 - 0,025.64 = 40,025 (g)
\(\Rightarrow C\%_{ZnSO_4}=\dfrac{0,025.161}{40,025}.100\%\approx10,056\%\)
a) \(n_{CuSO_4}=\dfrac{100.3,2\%}{160}=0,02\left(mol\right)\)
PTHH: CuSO4 + Fe ---> FeSO4 + Cu
0,02---->0,02--->0,02----->0,02
=> mFe (pư) = 0,02.56 = 1,12 (g)
b) mdd sau pư = 100 + 1,12 - 0,02.64 = 99,84 (g)
=> \(C\%_{FeSO_4}=\dfrac{0,02.152}{99,84}.100\%=3,045\%\)
a) \(n_{AgNO_3}=\dfrac{170.10\%}{170}=0,1\left(mol\right)\)
PTHH: Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag
0,05<--0,1--------->0,05--------->0,1
=> mCu (pư) = 0,05.64 = 3,2 (g)
b) mdd sau pư = 170 + 3,2 - 0,1.108 = 162,4 (g)
=> \(C\%_{Cu\left(NO_3\right)_2}=\dfrac{0,05.188}{162,4}.100\%=5,79\%\)
\(a,Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\\ n_{AgNO_3}=\dfrac{170.10\%}{170}=0,1\left(mol\right)=n_{Ag}\\ n_{Cu}=n_{Cu\left(NO_3\right)_2}=n_{AgNO_3}:2=0,1:2=0,05\left(mol\right)\\ m_{Cu}=0,05.64=3,2\left(g\right)\\ b,m_{ddsau}=m_{Cu}+m_{ddAgNO_3}-m_{Ag}=3,2+170-0,1.108=162,4\left(g\right)\\ C\%_{ddCu\left(NO_3\right)_2}=\dfrac{188.0,05}{162,4}.100\approx5,788\%\)
\(m_{ZnSO_4}=\dfrac{241,5.10}{100}=24,15\left(g\right)=>n_{ZnSO_4}=\dfrac{24,15}{161}=0,15\left(mol\right)\)
PTHH: 2Al + 3ZnSO4 --> Al2(SO4)3 + 3Zn
_____0,1<----0,15-------->0,05----->0,15
=> mAl = 0,1.27 = 2,7(g)
=> mZn = 0,15.65=9,75(g)
b) mdd sau pư = 2,7 + 241,5 - 9,75 = 234,45(g)
=> \(C\%\left(Al_2\left(SO_4\right)_3\right)=\dfrac{0,05.342}{234,45}.100\%=7,294\%\)
Ta có: \(n_{CuSO_4}=0,3\left(mol\right)\)
a, PT: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
______0,2____0,3_________________0,3 (mol)
b, \(m_{Al}=0,2.27=5,4\left(g\right)\)
c, \(m_{Cu}=0,3.64=19,2\left(g\right)\)
Bạn tham khảo nhé!
$n_{AgNO_3} = \dfrac{150.6,8\%}{170} =0,06(mol)$
$Cu+ 2AgNO_3 \to Cu(NO_3)_2 + 2Ag$
Theo PTHH :
$n_{Cu} = \dfrac{1}{2}n_{AgNO_3} = 0,03(mol)$
$m_{Cu} =0,03.64 = 1,92(gam)$
$n_{Ag} = n_{AgNO_3} = 0,06(mol)$
$\Rightarrow m_{dd\ sau\ pư} = 1,92 + 150 - 0,06.108 = 145,44(gam)$
$C\%_{Cu(NO_3)_2} = \dfrac{0,03.188}{145,44}.100\% = 3,88\%$