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a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b, \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
Theo PT: \(n_{CuO}=n_{CuCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,05.80=4\left(g\right)\)
c, \(C_{M_{CuCl_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
PTHH :
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05 0,1 0,05
\(b,m_{CuO}=0,05.80=4\left(g\right)\)
\(c,C_{M\left(CuCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
nAl= 0,5(mol)
a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2
nHCl= 6/2 . 0,5= 1,5(mol)
=>mHCl= 1,5.36,5=54,75(mol)
=> mddHCl= (54,75.100)/18,25=300(g)
b) nH2= 3/2. 0,5=0,75(mol)
=>V(H2,đktc)=0,75.22,4=16,8(l)
c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)
mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)
=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
0,1 0,2 0,1
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(C_{ddHCl}=\dfrac{7,3.100}{200}=3,65\)0/0
c) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)
Chúc bạn học tốt
a) \(m_{HCl}=200.7,3\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PTHH: CuO + 2HCl → CuCl2 + H2O
Mol: 0,2 0,4 0,2
b,\(m_{CuO}=0,2.80=16\left(g\right)\)
c, mdd sau pứ = 16+200 = 216 (g)
\(C\%_{ddCuCl_2}=\dfrac{0,2.135.100\%}{216}=12,5\%\)
mKOH=28(g)
nKOH=0.5(mol)
PTHH:2KOH+H2SO4->K2SO4+2H2O
a)Theo pthh:nH2SO4=1/2 nKOH->nH2SO4=0.25(mol)
mH2SO4=0.25*98=24.5(g)
C%ddH2SO4=24.5/100*100=24.5%
theo pthh:nK2SO4=nH2SO4->nK2SO4=0.25(mol)
mK2SO4=0.25*(39*2+96)=43.5(g)
c)mdd sau phản ứng:200+100=300(g)
d) C% muối=43.5:300*100=14.5%
\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(0.02.......0.02.................0.02\)
\(m_{H_2SO_4}=0.02\cdot98=1.96\left(g\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{1.96}{20\%}=9.8\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng }}=1.6+9.8=11.4\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{11.4}=28.07\%\)
Gọi nNa2CO3 = x (mol)
Na2CO3 + 2HCl \(\rightarrow\) 2NaCl + H2O + CO2
x \(\rightarrow\) 2x \(\rightarrow\) 2 x (mol)
C%(NaCl) = \(\frac{2.58,5x}{200+120}\) . 100% = 20%
=> x =0,547 (mol)
mNa2CO3 = 0,547 . 106 = 57,982 (g)
mHCl = 2 . 0,547 . 36,5 =39,931 (g)
C%(Na2CO3) =\(\frac{57,892}{200}\) . 100% = 28,946%
C%(HCl) = \(\frac{39,931}{120}\) . 100% = 33,28%
\(n_{CuO}=\dfrac{32}{80}=0,4(mol)\\ CuO+2HCl\to CuCl_2+H_2\\ \Rightarrow n_{HCl}=0,8(mol);n_{CuCl_2}=n_{H_2}=0,4(mol)\\ a,m_{dd_{HCl}}=\dfrac{0,8.36,5}{20\%}=146(g)\\ b,m_{CuCl_2}=0,4.135=54(g)\\ c,C\%_{CuCl_2}=\dfrac{54}{32+146-0,4.2}.100\%=30,47\%\)
\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)
\(n_{CuO}= \dfrac{32}{80}= 0,4 mol\)
Theo PTHH:
\(n_{HCl}= 2n_{CuO}= 0,8 mol\)
\(\Rightarrow m_{HCl}= 0,8 . 36,5=29,2 g\)
\(\rightarrow m_{dd HCl}= \dfrac{29,2 . 100%}{20%}= 146 g\)
b) Muối tạo thành là CuCl2
Theo PTHH:
\(n_{CuCl_2}= n_{CuO}= 0,4 mol\)
\(\Rightarrow m_{CuCl_2}= 0,4 . 135= 54g\)
c)
\(m_{dd sau pư}= m_{CuO} + m_{dd HCl}= 32 + 146=178 g\)
C%= \(\dfrac{54}{178} . 100\)%= 30,337 %