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\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,15 0,15 0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\
C_M=\dfrac{0,15}{0,1}=1,5M\)
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,5 0,5 0,5
\(m_{MgSO_4}=0,5.120=60g\\
V_{H_2}=0,5.22,4=11,2\left(mol\right)\\
\)
c)
\(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\\
pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\
LTL:0,5>0,2\)
=> H2SO4 dư
\(n_{Zn\left(p\text{ư}\right)}=n_{H_2SO_4}=0,2\left(mol\right)\\
n_{Zn\left(d\right)}=0,5-0,2=0,3\left(mol\right)\)
a) \(n_{H_2SO_4}=\dfrac{5,88}{98}=0,06\left(mol\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,04<--0,06------->0,02---------->0,06
\(\Rightarrow\left\{{}\begin{matrix}a=m_{Al}=0,04.27=1,08\left(g\right)\\V=V_{H_2}=0,06.22,4=1,344\left(l\right)\end{matrix}\right.\)
b)
Cách 1: \(m=m_{Al_2\left(SO_4\right)_3}=0,02.342=6,84\left(g\right)\)
Cách 2: \(m_{H_2}=0,06.2=0,12\left(g\right)\)
Áp dụng ĐLBTKL:
\(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)
\(\Rightarrow m=m_{Al_2\left(SO_4\right)_3}=1,08+5,88-0,12=6,84\left(g\right)\)
c) \(n_{O_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
PTHH: \(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,06}{2}< \dfrac{0,06}{1}\Rightarrow\) O2 dư, H2 hết
Theo PTHH: \(n_{O_2\left(p\text{ư}\right)}=\dfrac{1}{2}.n_{H_2}=0,03\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(d\text{ư}\right)}=0,06-0,03=0,03\left(mol\right)\\m_{O_2\left(d\text{ư}\right)}=0,03.32=0,96\left(g\right)\\V_{O_2\left(d\text{ư}\right)}=0,03.22,4=0,672\left(l\right)\end{matrix}\right.\)
Theo PTHH: \(n_{H_2O}=n_{H_2}=0,06\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,06.18=1,08\left(g\right)\)
Sửa đề : 11.2 g sắt
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(0.2....0.2.................0.2\)
\(m_{FeSO_4}=0.2\cdot152=30.4\left(g\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0.2}{0.05}=4\left(M\right)\)
Câu 1 :
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0.1.....................................0.1\)
\(m_{Zn}=0.1\cdot65=6.5\left(g\right)\)
Câu 2 :
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(n_{HCl}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(1..............2\)
\(0.2...........0.3\)
\(LTL:\dfrac{0.2}{1}>\dfrac{0.3}{2}\Rightarrow Fedư\)
\(m_{Fe\left(dư\right)}=\left(0.2-0.15\right)\cdot56=2.8\left(g\right)\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
a. PTHH:
Fe + 2HCl \(\rightarrow\)FeCl2 + H2
Mg + 2HCl \(\rightarrow\) MgCl2 + H2
Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
b.
Áp dụng định luật bảo toàn khổi lượng ta có:
mA + mB = mC + mD
m = mC + mD - mB
m = 42,6 + 0,7 - 25,55
m = 17,75
Vậy khối lượng của X bằng 17,75 g
Pt: Fe + H2SO4 --> FeSO4 + H2
.....Mg + H2SO4 --> MgSO4 + H2
.....Zn + H2SO4 --> ZnSO4 + H2
Theo pt: nH2SO4 = nH2 = 0,06 mol
=> nSO4 = nH2SO4 = 0,06 mol
a = mkim loại + mSO4 = 3,22 + 0,06 . 96 = 8,98 (g)
Pt: Fe + 2HCl --> FeCl2 + H2
......Mg + 2HCl --> MgCl2 + H2
......Zn + 2HCl --> ZnCl2 + H2
Theo pt: nHCl = 2nH2 = 2 . 0,06 = 0,12 mol
=> nCl - = 0,12 mol
a = mkim loại + mgốc axit = 3,22 + 0,12 . 36,5 = 7,6 (g)