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Câu 2
\(n_{H_2SO_4}=\dfrac{200.9,8\%}{98.100\%}=0,2\left(mol\right)\)
\(n_{KOH}=\dfrac{200.5,6\%}{56.100\%}=0,2\left(mol\right)\)
\(H_2SO_4+2KOH-->K_2SO_4+2H_2O\)
Ta có \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) => H2SO4 là chât còn dư
\(m_{K_2SO_4}=0,1.174=17,4\%\)
\(C\%_{K_2SO_4}=\dfrac{17,4}{200+200}.100\%=4,35\%\)
\(m_{H_2SO_4du}=\left(0,2-0,1\right).98=9,8\left(g\right)\)
\(C\%_{H_2SO_4du}=\dfrac{9,8}{200+200}.100\%=2,45\%\)
Đổi:400ml=0,4l
Gọi x;2y là số mol Fe,Al
Theo gt:\(m_{hhKL}\)=\(m_{Fe}+m_{Al}\)=56x+27y.2
=56x+54y=11(1)
Ta có PTHH:
Fe+\(H_2SO_4\)->\(FeSO_4\)+\(H_2\)(1)
x..........x................x.................(mol)
4Al+6\(H_2SO_4\)->2\(Al_2(SO_4)_3\)+3\(H_2\)(2)
2y...........3y...............y.........................(mol)
Ta có:\(C_{MddH_2SO_4}\)=1M
=>\(n_{H_2SO_4}\)=1.0,4=0,4mol
Theo PTHH(1);(2):
\(n_{H_2SO_4}\)=x+3y=0,4(2)
Từ (1);(2)=>\(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m_{Al}=54y=54.0,1=5,4\left(g\right)\\m_{Fe}=56x=56.0,1=5,6\left(g\right)\end{matrix}\right.\)
Theo PTHH(1);(2):\(n_{FeSO_4}\)=x=0,1(mol)
\(n_{Al_2\left(SO_4\right)_3}\)=y=0,1(mol)
Vậy \(C_{M\left(FeSO_4\right)}\)=0,1:0,4=0,25M
\(C_{MAl_2\left(SO_4\right)_3}\)=0,1:0,4=0,25M
\(n_{H_2SO_4}=0,4.1=0,4\left(mol\right)\)
Gọi x, y lần lượt là số mol của Al, Fe
Pt: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)
x \(\rightarrow\dfrac{3x}{2}\) \(\rightarrow0,1mol\)
Pt: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (2)
y \(\rightarrow y\) \(\rightarrow0,1mol\)
(1)(2) \(\Rightarrow\left\{{}\begin{matrix}1,5x+y=0,4\\27x+56y=11\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,4}=0,25M\)
\(C_{M_{FeSO_4}}=\dfrac{0,1}{0,4}=0,25M\)
nH2=0,15 mol
2Al+3H2SO4=>Al2(SO4)3+3H2
0,1 mol<= 0,15 mol
mAl=0,1.27=2,7g
nAl2(SO4)3=0,05 mol
=>mAl2(SO4)3=342.0,05=17,1g
nH2SO4=0,15 mol=>mH2SO4=14,7
mdd H2SO4=14,7/10%=147g
mdd sau pứ=2,7+147-0,15.2=149,4g
C%dd Al2(SO4)3=17,1/149,4.100%=11,45%
bài 2
\(n_{HCl}=0,4v_1\left(mol\right)\)
\(n_{NaOH}=v_2\left(mol\right)\)
\(n_{NaOHdu}=0,4.0,4=0,16\left(mol\right)\)
\(HCl+NaOH-->NaCl+H_2O\)
\(\left\{{}\begin{matrix}v_2-0,4v_1=0,16\\v_1+v_2=0,4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v_2\approx0,23\left(l\right)\\v_1\approx0,17\left(l\right)\end{matrix}\right.\)
Câu 1:
CuO + H2SO4 → CuSO4 + H2O
\(n_{CuO}=\frac{3,2}{80}=0,04\left(mol\right)\)
\(m_{H_2SO_4}=200\times9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\frac{19,6}{98}=0,2\left(mol\right)\)
Theo PT: \(n_{CuO}=n_{H_2SO_4}\)
Theo bài: \(n_{CuO}=\frac{1}{5}n_{H_2SO_4}\)
Vì \(\frac{1}{5}< 1\) ⇒ H2SO4 dư
Dung dịch sau pư gồm: H2SO4 dư và CuSO4
Ta có: \(m_{dd}saupư=3,2+200=203,2\left(g\right)\)
Theo Pt: \(n_{H_2SO_4}pư=n_{CuO}=0,04\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}dư=0,2-0,04=0,16\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}dư=0,16\times98=15,68\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}dư=\frac{15,68}{203,2}\times100\%=7,72\%\)
Theo Pt: \(n_{CuSO_4}=n_{CuO}=0,04\left(mol\right)\)
\(\Rightarrow m_{CuSO_4}=0,04\times160=6,4\left(g\right)\)
\(\Rightarrow C\%_{CuSO_4}=\frac{6,4}{203,2}\times100\%=3,15\%\)
Câu 2:
ZnO + H2SO4 → ZnSO4 + H2O
\(n_{ZnO}=\frac{8,1}{81}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=200\times24,5\%=49\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\frac{49}{98}=0,5\left(mol\right)\)
Theo Pt: \(n_{ZnO}=n_{H_2SO_4}\)
Theo bài: \(n_{ZnO}=\frac{1}{5}n_{H_2SO_4}\)
Vì \(\frac{1}{5}< 1\) ⇒ H2SO4 dư
Dung dịch sau pư gồm: H2SO4 dư và ZnSO4
Ta có: \(m_{dd}saupư=8,1+200=208,1\left(g\right)\)
Theo PT: \(n_{H_2SO_4}pư=n_{ZnO}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}dư=0,5-0,1=0,4\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,4\times98=39,2\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\frac{39,2}{208,1}\times100\%=18,84\%\)
Theo pT: \(n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\)
\(\Rightarrow m_{ZnSO_4}=0,1\times161=16,1\left(g\right)\)
\(\Rightarrow C\%_{ZnSO_4}=\frac{16,1}{208,1}\times100\%=7,74\%\)
đổi 100ml=0,1(l)
200ml=0,2(l)
ta có nKOH= 0.1x2= 0,2 (mol)
2KOH + H2SO4 -> K2SO4 + 2H2O
theo PTHH ta có nKOH=1/2nH2SO4=1/2nK2SO4
=> CM H2SO4 = 0.1/0.2= 0.5M
Ta có nK2SO4 = 0,1 mol =>V K2SO4 = 2,24(l)
=> CM K2SO4= 0.1/2,24\(\simeq\)0.45M