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$a)$
$Mg+H_2SO_4\to MgSO_4+H_2$
$b)$
$n_{Mg}=\frac{2,4}{24}=0,1(mol)$
Theo PT: $n_{MgSO_4}=n_{Mg}=0,1(mol)$
$\to m_{MgSO_4}=0,1.120=12(g)$
$c)$
$CuO+H_2\xrightarrow{t^o}Cu+H_2O$
Theo PT: $n_{Cu}=n_{H_2}=n_{Mg}=0,1(mol)$
$\to m_{Cu}=0,1.64=6,4(g)$
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)
c, Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,35}{1}\), ta được H2 dư.
Theo PT: \(n_{H_2\left(pư\right)}=n_{CuO}=0,15\left(mol\right)\)
\(\Rightarrow n_{H_2\left(dư\right)}=0,35-0,15=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2\left(dư\right)}=0,2.2=0,4\left(g\right)\)
`Zn+H_2SO_4->ZnSO_4+H_2`(to)
0,45-------------------0,45------0,45mol
`n_(Zn)=(29,25)/65=0,45mol`
`m_(ZnSO_4)=0,45.161=72,45g`
`V_(H_2)=0,45.22,4=10,08l`
c) `H_2+CuO->Cu+H_2O`(to)
0,45--------0,45 mol
`n_(Cu)=40/80=0,5 mol`
=>Cu dư , 0,05 mol
`m_(chất rắn)=0,45.64+0,05.80=32,8g`
\(n_{Zn}=\dfrac{m}{M}=\dfrac{29,25}{65}=0,45\left(mol\right)\)
a) \(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
1 1 1 1
0,45 0,45 0,45 0,45
b) \(m_{ZnSO_4}=n.M=0,45.\left(65+32+16.4\right)=51,03\left(g\right)\\ V_{H_2}=n.24,79=0,45.24,79=11,1555\left(l\right)\)
c) \(n_{CuO}=\dfrac{m}{M}=\dfrac{40}{\left(64+16\right)}=0,5\left(mol\right)\)
\(PTHH:CuO+H_2\rightarrow Cu+H_2O\)
1 1 1 1
0,5 0,5 0,5 0,5
\(m_{Cu}=0,5.64=32\left(g\right).\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
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
\(a,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ PTHH:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\left(1\right)\\ Theo.pt\left(1\right):n_{H_2}=n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\\ V_{H_2}=0,1.22,4=2,24\left(l\right)\\ m_{MgSO_4}=0,1.120=12\left(g\right)\\ b,PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\left(2\right)\\ Theo.pt\left(2\right):n_{Cu}=n_{H_2}=0,1\left(mol\right)\\ m_{Cu}=0,1.64=6,4\left(g\right)\)
a)
\(Mg + H_2SO_4 \to MgSO_4 + H_2\\ n_{H_2} = n_{Mg} = \dfrac{3,6}{24} = 0,15(mol)\\ b)\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{H_2} = 0,15(mol)\\ \Rightarrow m_{Cu} = 0,15.64 = 9,6(gam)\)
a)
\(Mg + H_2SO_4 \to MgSO_4 + H_2\)
Magie tan dần, xuất hiện bọt khí không màu không mùi.
b)
\(n_{H_2} = n_{MgSO_4} = n_{Mg} = \dfrac{9,6}{24} = 0,4(mol)\\ m_{MgSO_4} = 0,4.120 = 48(gam)\\ V_{H_2} = 0,4.22,4 = 8,96(lít)\)
c)
\(CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{H_2} = 0,4(mol)\\ \Rightarrow m_{Cu} = 0,4.64 = 25,6(gam)\)