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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
PTHH: MgO + 2HCl ---> MgCl2 + H2.
Theo PT: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)
=> \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)
Ta có: \(m_{dd_{MgCl_2}}=4+100=104\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{9,5}{104}.100\%=9,13\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)
\(n_{CuO} = \dfrac{8}{80}= 0,1 mol\)
\(m_{HCl}= 100 . 18,25\)%= 18,25g
\(n_{HCl}= \dfrac{18,25}{36,5}=0,5 mol\)
\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)
Trước pư: 0,1 0,5
PƯ 0,1 0,2 0,1
Sau pư 0 0,3 0,1
Dung dịch sau pư: CuCl2 và HCl dư
\(m_{dd sau pư} = m_{CuO} + m_{dd HCl}=8 +100= 108g\)
C%HCl\(=\dfrac{0,3 . 36,5}{108} . 100\)%= 10,139%
C%CuCl2\(=\dfrac{0,1. 135}{108}.100\)%= 12,5%
![](https://rs.olm.vn/images/avt/0.png?1311)
a. Ta có: \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)
Ta lại có: \(C_{\%_{HCl}}=\dfrac{m_{ct_{HCl}}}{100}.100\%=7,3\%\)
=> mHCl = 7,3(g)
=> \(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
PTHH:
Fe3O4 + 8HCl ---> FeCl2 + 2FeCl3 + 4H2O
1 ---> 8
0,1 ---> 0,2
=> \(\dfrac{0,1}{1}>\dfrac{0,2}{8}\)
Vậy Fe3O4 dư
=> mdư = 23,2 - 7,3 = 15,9 (g)
b. Theo PT: \(n_{FeCl_2}=\dfrac{1}{8}.n_{HCl}=\dfrac{1}{8}.0,2=0,025\left(mol\right)\)
=> \(m_{FeCl_2}=0,025.127=3,175\left(g\right)\)
Theo PT: \(n_{FeCl_3}=\dfrac{1}{4}.n_{HCl}=\dfrac{1}{4}.0,2=0,05\left(mol\right)\)
=> \(m_{FeCl_3}=0,05.162,5=8,125\left(g\right)\)
=> \(m_{muối}=8,125+3,175=11,3\left(g\right)\)
c. Ta có: mdung dịch sau PỨ = \(23,2+100=123,2\left(g\right)\)
Theo PT: \(n_{H_2O}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
=> \(m_{H_2O}=0,1.18=1,8\left(g\right)\)
mcác chất sau PỨ = 1,8 + 11,3 = 13,1(g)
=> \(C_{\%_{sauPỨ}}=\dfrac{13,1}{123,2}.100\%=10,63\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=0,204\left(mol\right)\)
PTHH:
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
0,04 0,04 0,04
\(\dfrac{0,04}{1}< \dfrac{0,204}{1}\) --> H2SO4 dư
\(C\%_{CuSO_4}=\dfrac{0,04.160}{3,2+100}.100\%=6,2\%\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{3,2+100}.100\%=19\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)
PTHH: CuO + H2SO4 → CuSO4 + H2O
Mol: 0,02 0,02 0,02
Ta có: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) ⇒ CuO hết, H2SO4 dư
mdd sau pứ = 1,6 + 100 = 101,6 (g)
\(C\%_{ddCuSO_4}=\dfrac{0,02.160.100\%}{101,6}=3,15\%\)
\(C\%_{ddH_2SO_4}=\dfrac{\left(\dfrac{10}{49}-0,02\right).98.100\%}{101,6}=17,76\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ncuo= 1,6/80=0,02
nh2so4=(100*20)/( 98*100)= 0,2> 0,02-> cuo pư hết, h2so4 dư
cuo+ h2so4-> cuso4+h2o
0,02-> 0,02 0,02
mdd sau pư= 1,6+ 100= 101,6
c%h2so4 dư= (0,2-0,02)*98/101,6*100= 17,36%
c%cuso4= 0,02*160/101,6*100= 3,15%
![](https://rs.olm.vn/images/avt/0.png?1311)
nCuO= \(\frac{1,6}{80}\) = 0,02 (mol)
\(n_{H_2SO_4}\) = \(\frac{100.20\%}{98}\) =0,2041(mol)
CuO + H2SO4 \(\rightarrow\) CuSO4 + H2O
bđ 0,02 \(\frac{10}{49}\) (mol)
pư 0,02 \(\rightarrow\) 0,02 \(\rightarrow\) 0,02 (mol)
spư 0 0,1841 0,02 (mol)
md d (sau pư) = 100 + 1,6 = 101,6 (g)
C%(CuSO4) = \(\frac{0,02.160}{101,6}\) . 100% = 3,15%
C%(H2SO4)= \(\frac{0,1841.98}{101,6}\) . 100% = 17,76%
Cho 4g CuO tác dụng với 100g dd HCl 18,25%. Tính nồng độ % các chất có trong dd sau phản ứng
---
nCuO= 4/80= 0,05(MOL)
mHCl= 18,25% . 100= 18,25(g)
=> nHCl= 18,25/36,5= 0,5(mol)
PTHH: CuO + 2 HCl -> CuCl2 + H2O
0,05______0,1__________0,05(mol)
Ta có: 0,05/1 < 0,5/2
=> HCl dư, CuO hết, tính theo nCuO
mHCl(dư)= (0,5 - 0,05.2).36,5=14,6(g)
mCuCl2= 0,05.135= 6,75(g)
mddsau= mddHCl + mCuO= 100 +4=104(g)
=> \(C\%ddHCl\left(dư\right)=\frac{14,6}{104}.100\approx14,038\%\\ C\%ddCuCl2=\frac{6,75}{104}.100\approx6,49\%\)
\(n_{CuO}=\frac{4}{80}=0,05\left(mol\right)\)
\(n_{HCl}=\frac{100.18,25\%}{36,5}=0,5\left(mol\right)\)
\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\)
Ban đầu_ 0,05 ___0,5
Phứng__0,05______0,1
Sau ____0 _____0,4______0,05 ____________
\(C\%_{HCl\left(Dư\right)}=\frac{0,4.36,5}{4+100}.100\%=14,04\%\)
\(C\%_{CuCl2}=\frac{0,05.135}{4+100}.100\%=6,49\%\)