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\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH :
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,1 0,1 0,1 0,1
\(a,m_{MgSO_4}=0,1.120=12\left(g\right)\)
\(b,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(c,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{10}=98\left(g\right)\)
\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{4.8}{24}=0,2mol\)
đổi 200 ml = 0,2 l
PTHH: Mg + 2HCl \(\rightarrow\) MgCl2 + H2
TL; 1 2 1 1
mol: 0,2 \(\rightarrow\) 0,2 \(\rightarrow\) 0,2
b. \(C_{M_{ddHCl}}=\dfrac{n_{HCl}}{V_{dd_{HCl}}}=\dfrac{0,2}{0,2}=1M\)
\(c.V_{H_2}=n_{H_2}.22,4=0,2.22,4=4,48l\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
a) PTHH : \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b) \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow C_{MddHCl}=\dfrac{0,4}{0,2}=2M\)
c) \(n_{Mg}=n_{H2}=0,2\left(mol\right)\Rightarrow V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
a)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2--->0,3------->0,1------------>0,3
$m_{dd.H_2SO_4}=\frac{0,3.98.100\%}{19,6\%}=150\left(g\right)$
b)
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342.100\%}{5,4+150-0,3.2}=22,09\%\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
_____0,2_______0,3________0,1_______0,3 (mol)
a, \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{19,6\%}=150\left(g\right)\)
b, \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, Ta có: m dd sau pư = 5,4 + 150 - 0,3.2 = 154,8 (g)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{154,8}.100\%\approx22,09\%\)
a)
$n_{Al} = 0,3(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH :
$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$
b)
$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$
c)
$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$
Bài 4 :
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)0/0
d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)
\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)0/0
Chúc bạn học tốt
Bài 3 :
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,5 0,5 0,5 0,5
b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)
c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)0/0
d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)
\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)
\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)0/0
Chúc bạn học tốt
a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
PTHH: 2CH3COOH + Mg --> (CH3COO)2Mg + H2
0,1<----------------------0,05------->0,05
=> VH2 = 0,05.22,4 = 1,12 (l)
b) \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,1}{0,025}=4M\)
PTHH: \(Mg+H_2SO_4\rightarrow MgsO_4+H_2\uparrow\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)=n_{MgSO_4}=n_{H_2SO_4}\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgSO_4}=0,2\cdot120=24\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\C\%_{H_2SO_4}=\dfrac{0,2\cdot98}{294}\cdot100\%\approx6,67\%\end{matrix}\right.\)
Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)
a. PTHH: Mg + H2SO4 ---> MgSO4 + H2↑
Theo PT: \(n_{H_2}=n_{Mg}=0,4\left(mol\right)\)
=> \(V_{H_2}=0,4.22,4=8,96\left(lít\right)\)
b. Theo PT: \(n_{H_2SO_4}=n_{Mg}=0,4\left(mol\right)\)
=> \(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{39,2}{m_{dd_{H_2SO_4}}}.100\%=10\%\)
=> \(m_{dd_{H_2SO_4}}=392\left(g\right)\)
c. Ta có: \(m_{H_2}=0,4.2=0,8\left(g\right)\)
=> \(m_{dd_{MgSO_4}}=9,6+392-0,8=400,8\left(g\right)\)
Theo PT: \(n_{MgSO_4}=n_{Mg}=0,4\left(mol\right)\)
=> \(m_{MgSO_4}=0,4.120=48\left(g\right)\)
=> \(C_{\%_{MgSO_4}}=\dfrac{48}{400,8}.100\%=11,98\%\)