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a: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b: \(n_{H2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(\Leftrightarrow n_{Al}=0.1\left(mol\right)\)
\(m_{Al}=n_{Al}\cdot M_{Al}=0.1\cdot27=2.7\left(g\right)\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2--->0,3-------->0,1----------->0,3
b) `V_{H_2} = 0,3.22,4 = 6,72 (l)`
c) `m_{H_2SO_4} = 0,3.98 = 29,4 9g)`
d) \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Xét tỉ lệ: 0,2 < 0,3 => H2 dư
a) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
b) nAl = \(\frac{40,5}{27}=1,5\left(mol\right)\)
Từ PT \(\Rightarrow n_{H_2SO_4}=2,25\left(mol\right);n_{Al_2\left(SO_4\right)_3}=0,75\left(mol\right);n_{H_2}=2,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=2,25.98=220,5\left(g\right)\)
c) \(m_{Al_2\left(SO_4\right)_3}=0,75.342=256,5\left(g\right)\)
d) đktc : \(V_{H_2}=22,4.2,25=50,4\left(l\right)\)
a) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2 (1)
b) nAl = 40,5 : 27 = 1,5 mol
Từ pt(1) suy ra : nH2SO4 = \(\frac{3}{2}nAl\) = \(\frac{3}{2}.1,5=2,25mol\)
Khối lượng H2SO4 là : mH2SO4 = 2,25 . 98 = 220,5 g
c) Từ pt(1) => nAl2(SO4)3 = \(\frac{1}{2}nAl=\frac{1}{2}.1,5=0,75mol\)
=> mAl2(SO4)3 = 0,75 . 342 = 256,5 g
d) Từ pt(1) => nH2 = nH2SO4 = 2,25 mol
Thể tích khí H2 là : VH2=2,25 . 22,4 = 50,4 lit
2Al+3H2SO4->Al2(SO4)3+3H2
0,1----------------------0,075----0,15
n H2=0,15 mol
=>mAl=0,1.27=2,7g
=>m Al2(SO4)3=0,075.342=25,65g
a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)
\(m_{Al}=0,1.27=2,7\left(g\right)\)
c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
b, Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)
0,2<--0,6<----------0,2<------0,3 (mol)
\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(m_{HCl}=n\cdot M=0,6\cdot\left(1+35,5\right)=21,9\left(g\right)\)
\(m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)
a, PT: 2Al+6HCl→2AlCl3+3H2
Ta có: nH2=6,7222,4=0,3(mol)
Theo PT: nHCl=2nH2=0,6(mol)
⇒mHCl=0,6.36,5=21,9(g)
b, Theo PT: nAl=23nH2=0,2(mol)
⇒mAl=0,2.27=5,4(g)
Câu 7 :
\(n_{H2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)
Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)
2 3 1 3
\(\dfrac{8}{15}\) 0,8 \(\dfrac{4}{15}\) 0,8
\(n_{H2SO4}=\dfrac{0,8.3}{3}=0,8\left(mol\right)\)
⇒ \(m_{H2SO4}=0,8.98=78,4\left(g\right)\)
\(n_{Al2\left(SO4\right)3}=\dfrac{0,8.1}{3}=\dfrac{4}{15}\left(mol\right)\)
⇒ \(m_{Al2\left(SO4\right)3}=\dfrac{4}{15}.342=91,2\left(g\right)\)
\(n_{Al}=\dfrac{0,8.2}{3}=\dfrac{8}{15}\left(mol\right)\)
⇒ \(m_{Al}=\dfrac{8}{15}.27=14,4\left(g\right)\)
Chúc bạn học tốt
bài 8
Al2O3+6HNO3->2Al(NO3)3+3H2
\(\dfrac{2}{15}\)------------0,8-------\(\dfrac{4}{15}\)----------0,4 mol
n H2O=\(\dfrac{7,2}{18}\)=0,4 mol
=>m Al2O3=\(\dfrac{2}{15}\).102=13,6g
=>m HNO3=0,8.63=50,4g
=>m Al(NO3)3=\(\dfrac{4}{15}\).213=56,8g