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a,\(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right);n_{H_2SO_4}=1,5.0,2=0,3\left(mol\right)\)
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,3 0,3 0,3
Ta có: \(\dfrac{0,4}{1}>\dfrac{0,3}{1}\) ⇒ Mg dư, H2SO4 pứ hết
\(m_{MgSO_4}=0,3.120=36\left(g\right)\)
b,\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{Fe_2O_3}=\dfrac{6,4}{160}=0,04\left(mol\right)\)
PTHH: 3H2 + Fe2O3 → 2Fe + 3H2O
Mol: 0,04 0,08
Ta có: \(\dfrac{0,3}{3}>\dfrac{0,04}{1}\) ⇒ H2 dư, Fe2O3 pứ hết
\(\Rightarrow m_{Fe}=0,08.56=4,48\left(g\right)\)
\(nMg=\dfrac{12}{24}=0,5\left(mol\right)\)
\(nH_2SO_4=\dfrac{29,4}{98}=0,3\left(mol\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
1 1 1 1 (mol)
0,3 0,3 0,3 0,3 (mol)
LTL : 0,5/1 > 0,3/1
=> Mg dư , H2SO4 đủ
\(VH_2=0,3.22,4=6,72\left(l\right)\)
m muối là mMgSO4
=> \(m\left(muối\right)=mMgSO_4=0,3.120=36\left(g\right)\)
a)
\(Mg + H_2SO_4 \to MgSO_4 + H_2\\ n_{H_2} = n_{Mg} = \dfrac{3,6}{24} = 0,15(mol)\\ b)\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{H_2} = 0,15(mol)\\ \Rightarrow m_{Cu} = 0,15.64 = 9,6(gam)\)
`Mg + H_2 SO_4 -> MgSO_4 + H_2`
`0,45` `0,45` `0,45` `(mol)`
`n_[H_2 SO_4] = [ [ 14,7 ] / 100 . 300 ] / 98 = 0,45 (mol)`
`a) V_[H_2] = 0,45 . 22,4 = 10,08 (l)`
`b) m_[MgSO_4] = 0,45 . 120 = 54 (g)`
\(n_{H_2SO_4}=\dfrac{300.14,7\%}{98}=0,45\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,45 0,225 0,225
\(V_{H_2}=0,225.22,4=5,04l\\
m_{MgCl_2}=95.0,225=21,375g\)
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) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
n Zn= 19,5/65=0,3 (mol).
PTPƯ: Zn(0.3) + HCl(0.6) ----> ZnCl2(0.3) + H2(0,3)
mHCl=0,6.36.5=21.9(g)
a) C%HCl= 21.9/300.100%=7,3%
b) VH2=0,3.22,4=6,72(lít)
c) mH2=0,3.2=0,6(g)
mZnCl2=0,3.136=40,8(g)
mddZnCl2 =(19,5+300)-0,6=318,9(g)
C%=mZnCl2/mddZnCl2.100= 40,8/318,9.100=12,793%
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeSO_4}=0,2.152=30,4\left(g\right)\)
c, \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
d, \(n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,4 0,8 0,4 0,4
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b,C\%_{HCl}=\dfrac{0,8.36,5}{150}.100\%=19,5\%\\ c,m_{\text{dd}}=26+150-\left(0,4.2\right)=175,2\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,4.136}{175,2}.100\%=31\%\)
nMg = 3,6 : 24 = 0,15 (mol)
pthh : Mg + 2HCl --> MgCl2 + H2
0,15-----------> 0,15 --->0,15 (mol)
mMgCl2 = 0,15 . 95 = 14,25 (mol)
VH2 (đkc)= 0,15. 24,79 = 3,718(l)
\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,15 0,3 0,15 0,15
\(m_{MgCl_2}=0,15\cdot95=14,25g\)
\(V_{H_2}=0,15\cdot22,4=3,36l\)
nMg = 4,8 : 24 = 0,2 mol
a) Mg + H2SO4 → MgSO4 + H2
Theo tỉ lệ phản ứng => nH2SO4 phản ứng = nMgSO4 = nH2 = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48 lít.
b)
mH2SO4 phản ứng = 0,2.98 = 19,6 gam
=> C% H2SO4 = \(\dfrac{19,6}{300}.100\text{%}\) = 6,53%
c) mMgSO4 = 0,2.120 = 24 gam.