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a, \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b, Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{4}{15}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{\dfrac{4}{15}.27}{10}.100\%=72\%\\\%m_{Cu}=28\%\end{matrix}\right.\)
c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{39,2}{300}.100\%\approx13,067\%\)
\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)
1 1 1 1
0,02 0,02
\(n_{CuSO4}=\dfrac{0,02.1}{1}=0,02\left(mol\right)\)
⇒ \(m_{CuSO4}=0,02,160=3,2\left(g\right)\)
\(m_{ddspu}=1,6+300=301,6\left(g\right)\)
\(C_{CuSO4}=\dfrac{3,2.100}{301,6}=1,6\)0/0
Chúc bạn học tốt
nNa=4,6/23=0,2(mol)
PTHH: Na + H2O -> NaOH + 1/2 H2
0,2_______________0,2____0,1(mol)
mddNaOH=4,6+100-0,1.2=104,4(g)
mNaOH=0,2.40=8(g)
=>C%ddNaOH= (8/104,4).100=7,663%
=> Chọn B (gần nhất)
\(n_K=\frac{5,85}{15}=0,15(mol)\\ K+H_2O \to KOH +\frac{1}{2}H_2\\ n_{KOH}=n_K=0,15(mol)\\ n_{H_2}=\frac{1}{2}.n_K=\frac{1}{2}.0,15=0,075(mol)\\ m_{dd}=5,85+100-(0,075.2)=105,7(g)\\ C\%=\frac{0,15.56}{105,7}.100=7,95\%\)
Bài 4 :
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)0/0
d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)
\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)0/0
Chúc bạn học tốt
Bài 3 :
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,5 0,5 0,5 0,5
b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)
c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)0/0
d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)
\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)
\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)0/0
Chúc bạn học tốt
\(a)CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ b)n_{CuO}=\dfrac{4}{80}=0,05mol\\ n_{H_2SO_4}=\dfrac{100.20}{100.98}=\dfrac{10}{49}mol\\ \Rightarrow\dfrac{0,05}{1}< \dfrac{10:49}{1}\rightarrow H_2SO_4.dư\\ n_{CuSO_4}=n_{H_2SO_4}=n_{CuO}=0,05mol\\ C_{\%CuSO_4}=\dfrac{0,05.160}{100+4}\cdot100=7,69\%\\ C_{\%H_2SO_4}=\dfrac{\left(10:49-0,05\right)98}{100+4}\cdot100=14,52\%\)
\(n_{Al}=0,3\left(mol\right);n_{H_2SO_4}=0,75\left(mol\right)\)
Dễ thấy \(H_2SO_4\) dư \(\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,15\left(mol\right);n_{H_2}=0,45\left(mol\right)\)
\(m_{\text{dd sau pư}}=100+8,1-0,45.2=107,2\left(g\right)\)
\(\Rightarrow C\%=\dfrac{0,15.342}{107,2}.100\%=47,85\%\)
Cái này đúng ra là \(C\%=47,85447761\%\approx47,86\%\)
A