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\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2____0,4_____________0,2
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(V_{H2}=0,2.22,4=4,48\left(l\right)\)
\(4H_2+Fe_3O_4\rightarrow3Fe+4H_2O\)
0,2_____________0,15____
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
Câu 1:
a, Thí nghiệm 1:
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Thí nghiệm 2:
\(3H_2+Fe_2O_3\underrightarrow{^{t^o}}2Fe+3H_2O\)
b, Ta có:
\(n_{Mg}=\frac{13,44}{24}=0,56\left(mol\right)\)
\(\Rightarrow n_{H2}=n_{MgSO4}=n_{Mg}=0,56\left(mol\right)\)
\(\Rightarrow V_{H2}=0,56.22,4=12,544\left(l\right)\)
\(\Rightarrow m_{MgSO4}=0,56.120=67,2\left(g\right)\)
c,\(n_{Fe2O3}=\frac{1}{3}n_{H2}=\frac{1}{3}.0,56=\frac{14}{75}\left(mol\right)\)
\(\Rightarrow m_{Fe2O3}=\frac{14}{75}.160=\frac{448}{15}\left(g\right)\)
Câu 2:
d,
\(Zn+2HCl\rightarrow ZnCl_2+H_2\left(1\right)\)
\(Fe_3O_4+4H_2\underrightarrow{^{t^o}}3Fe+4H_2O\left(2\right)\)
e, Ta có:
\(n_{H2}=\frac{5,376}{22,4}=0,24\left(mol\right)\)
Theo PTHH1:
\(n_{HCl}=2n_{H2}=0,48\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4.36,5=17,52\left(g\right)\)
\(n_{ZnCl2}=n_{H2}=0,24\left(mol\right)\)
\(\Rightarrow m_{ZnCl2}=0,24.136=32,64\left(g\right)\)
f, Theo PTHH2:
\(n_{Fe3O4}=\frac{1}{4}n_{H2}=0,06\left(mol\right)\)
\(\Rightarrow m_{Fe3O4}=0,06.232=13,92\left(g\right)\)
a, nZn = 26/65 = 0,4 (mol)
PTHH: Zn + 2HCl -> ZnCl2 + H2
nZn = nH2 = 0,4 (mol)
VH2 = 0,4 . 22,4 = 8,96 (l)
b, nFe2O3 = 16/160 = 0,1 (mol)
PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O
LTL: 0,1 < 0,4/3 => H2 dư
nFe = 0,1 . 3 = 0,3 (mol)
mFe = 0,3 . 56 = 16,8 (g)
a) \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,4--------------------->0,4
=> VH2 = 0,4.22,4 = 8,96 (l)
b)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\) => Fe2O3 hết, H2 dư
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,1---------------->0,2
=> mFe = 0,2.56 = 11,2 (g)
a, PTHH:
H2 + ZnO → Zn + H2O
nZnO = 8,1 / 81 = 0,1 ( mol)
Thep PTHH nH2 = nZnO = 0,1( mol)
nzn = nZnO = 0,1 (mol)
VH2 = 0,1 x 22,4 = 2,24 (l)
b, mZn = 0,1 x 65 = 6,5 (g)
c, Zn + 2HCl → ZnCl2 + H2
mHCl = 200 x 7,3 % = 14,6 ( g)
nHCl = 14,6 / 36,5 = 0,4 ( mol)
Theo PTHH nH2 = 1/2nHCl= 0,4 /2 = 0,2( mol)
VH2 = 0,2 x 22,4 = 4,48( l)
d, y H2 + FexOy → x Fe + yH2O
Theo câu a nH2 = 0,1 ( mol)
Theo PTHH nFexOy= 1/ynH2 = 0,1 /y ( mol)
mFexOy = 0,1/y( 56x + 16y)= 3,24 (g)
đoạn này bạn tự tính nhé!
a, \(PTHH:CuO+H_2\underrightarrow{^{t^o}}Cu+H_2O\)
Ta có:
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow n_{Cu}=n_{H2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)
b,
i .\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
ii. \(n_{FeCl2}=116,2\left(g\right)\) ( Đề cho)
iii: \(n_{FeCl2}=\frac{116,2}{127}=0,9\left(mol\right)\)
\(\Rightarrow n_{HCl}=2n_{FeCl2}=1,8\left(mol\right)\)
\(\Rightarrow m_{HCl}=1,8.36,5=65,7\left(g\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{HCl}=2n_{Zn}=0,4(mol)\\ \Rightarrow m_{HCl}=0,4.36,5=14,6(g)\\ c,n_{H_2}=n_{Zn}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\)
b) mHCl = 14,6 (g)
V H2 = 4,48 (l)
Giải thích các bước:
a) PTHH: Zn + 2HCl → ZnCl2 + H2↑
b) nZn = 13 : 65 = 0,2 mol
Theo PTHH: nHCl = 2.nZn = 0,4 mol
mHCl = 0,4 . 36,5 = 14,6(g)
c) nH2 = nZn = 0,2 mol
VH2 = 0,2 . 22,4 = 4,48 (l)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
+\(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
+\(nH_2=n_{Zn}=0,5\left(mol\right)\)
+\(n_{HCl}=2n_{Zn}=1\left(mol\right)\)
+\(V_{H2}=0,5.22,4=11,2\left(lit\right)\)
\(m_{HCl}=1.36,5=36,5\left(gam\right)\)
\(n_{Zn}=\dfrac{m}{M}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
\(Zn\) \(+\) \(2\)\(HCl\) → \(ZnCl_2\) \(+\) \(H_2\)
\(0,5\) \(mol\) → \(1\) \(mol\) → \(0,5\)\(mol\) → \(0,5\) \(mol\)
\(V_{H_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)
\(m_{HCl}=n.M=1.36,5=36,4\left(g\right)\)