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\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,1 0,1 0,1
\(m_{MgSO_4}=120.0,1=12\left(g\right)\\
n_{CuO}=\dfrac{48}{80}=0,6\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(LTL:\dfrac{0,6}{1}>\dfrac{0,1}{1}\)
=> CuO dư
\(n_{CuO\left(P\text{Ư}\right)}=n_{H_2}=0,1\left(mol\right)\\
m_{CuO\left(d\right)}=\left(0,6-0,1\right).80=40\left(g\right)\)
nMg=2,424=0,1(mol)pthh:Mg+H2SO4→MgSO4+H2nMg=2,424=0,1(mol)pthh:Mg+H2SO4→MgSO4+H2
0,1 0,1 0,1
mMgSO4=120.0,1=12(g)nCuO=4880=0,6(mol)pthh:CuO+H2to→Cu+H2OmMgSO4=120.0,1=12(g)nCuO=4880=0,6(mol)pthh:CuO+H2to→Cu+H2O
Bài 1:
\(a,PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ b,m_{Zn}+m_{H_2SO_4}=m_{ZnSO_4}+m_{H_2}\\ c,m_{H_2SO_4}=32,2+0,4-13=19,6(g) \)
Bài 2:
Bảo toàn KL: \(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)
\(\Rightarrow m_{H_2}=6,5+7,3-13,6=0,2(g)\)
Bài 3:
Bảo toàn KL: \(m_{Mg}+m_{O_2}=m_{MgO}\)
\(\Rightarrow m_{O_2}=1000-600=400(g)\)
a) PTHH: Fe + H2SO4 ===> FeSO4 + H2
b) Ta có: nFe =
Theo PTHH, nH2SO4 = nFe = 0,25 (mol)
=> mH2SO4 = 0,25 x 98 = 24,5 (gam)
c) Theo PTHH, nH2 = nFe = 0,25 (mol)
=> VH2(đktc) = 0,25 x 22,4 = 5,6 (l)
d) Theo PTHH, nFeSO4 = nFe = 0,25 (mol)
=> mFeSO4(tạo thành) = 0,25 x 152 = 38 (gam)
Phương trình hóa học: 2NaOH + H 2 S O 4 → N a 2 S O 4 + 2 H 2 O
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có: \(n_{NaOH}=\dfrac{14,8}{40}=0,37\left(mol\right)\)
a, Theo PT: \(n_{Na}=n_{NaOH}=0,37\left(mol\right)\)
\(\Rightarrow A_{Na}=0,37.6.10^{23}=2,22.10^{23}\) (nguyên tử)
\(m_{Na}=0,37.23=8,51\left(g\right)\)
b, \(n_{H_2}=\dfrac{1}{2}n_{NaOH}=0,185\left(mol\right)\)
\(\Rightarrow A_{H_2}=0,185.6.10^{23}=1,11.10^{23}\) (phân tử)
\(m_{H_2}=0,185.2=0,37\left(g\right)\)
c, \(V_{H_2}=0,185.24,79=4,58615\left(l\right)\)
\(a,n_{Al}=\dfrac{4,5}{27}=\dfrac{1}{6}\left(mol\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(\dfrac{1}{6}\)-->\(0,25\)-------->\(\dfrac{1}{12}\)------------>0,25
\(V_{ddH_2SO_4}=\dfrac{0,25}{1,5}=\dfrac{1}{6}\left(l\right)\\ b,m_{muối}=\dfrac{1}{12}.342=28,5\left(g\right)\\ V_{H_2}=0,25.22,4=5,6\left(l\right)\)
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
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{12,25}{98}=0,125mol\)
\(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\)
0,25 0,125 0,125 ( mol )
\(m_{Na}=n.M=0,25.23=5,75g\)
\(m_{Na_2SO_4}=0,125.142=17,75g\)