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\(a)Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=\dfrac{5,28}{24}=0,22mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,22...0,44........0,22........0,22\\ V_{H_2\left(đkc\right)}=0,22.24,79=5,4538l\\ b)C_{\%HCl}=\dfrac{0,44.26,5}{200}\cdot100=8,03\%\\ c)C_{\%MgCl_2}=\dfrac{0,22.95}{200+5,28}\cdot100\approx10,18\%\)
Bài 14:
Ta có: \(n_{BaCO_3}=\dfrac{39,4}{197}=0,2\left(mol\right)\)
PT: \(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\)
a, \(n_{CO_2}=n_{BaCO_3}=0,2\left(mol\right)\Rightarrow V_{CO_2}=0,2.24,79=4,958\left(l\right)\)
b, Sửa đề: tính khối lượng dung dịch HCl → tính nồng độ % dd HCl.
\(n_{HCl}=2n_{BaCO_3}=0,4\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{100}.100\%=14,6\%\)
c, \(n_{BaCl_2}=n_{BaCO_3}=0,2\left(mol\right)\)
Ta có: m dd sau pư = 39,4 + 100 - 0,2.44 = 130,6 (g)
\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,2.208}{130,6}.100\%\approx31,85\%\)
Bài 12:
Ta có: \(n_{MgCO_3}=\dfrac{25,2}{84}=0,3\left(mol\right)\)
PT: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
a, Theo PT: \(n_{CO_2}=n_{MgCO_3}=0,3\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,3.24,79=7,437\left(l\right)\)
b, Ta có: m dd sau pư = 25,2 + 200 - 0,3.44 = 212 (g)
Theo PT: \(n_{MgCl_2}=n_{MgCO_3}=0,3\left(mol\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,3.95}{212}.100\%\approx13,44\%\)
Bài 13:
Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
1. \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\) \(\Rightarrow V_{CO_2}=0,1.24,79=2,479\left(l\right)\)
2. \(n_{HCl}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{7,3\%}=100\left(g\right)\)
3. Ta có: m dd sau pư = 10 + 100 - 0,1.44 = 105,6 (g)
Theo PT: \(n_{CaCl_2}=n_{CaCO_3}=0,1\left(mol\right)\)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,1.111}{105,6}.100\%\approx10,51\%\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,2 0,4 0,2
\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6.100\%}{20\%}=73\left(g\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(25^oC,1bar\right)}=0,2.22,4=4,48\left(l\right)\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{HCl}=0,4.36,5=14,6\left(g\right)\)
nAl= 0,5(mol)
a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2
nHCl= 6/2 . 0,5= 1,5(mol)
=>mHCl= 1,5.36,5=54,75(mol)
=> mddHCl= (54,75.100)/18,25=300(g)
b) nH2= 3/2. 0,5=0,75(mol)
=>V(H2,đktc)=0,75.22,4=16,8(l)
c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)
mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)
=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)
a) $CaSO_3 + 2HCl \to CaCl_2 + SO_2 + H_2O$
b)
$n_{SO_2} = n_{CaSO_3} = \dfrac{12}{120} = 0,1(mol)$
$m_{SO_2} = 0,1.64 = 6,4(gam)$
c)
$n_{HCl} = 2n_{SO_2} = 0,2(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
d)
$m_{dd\ sau\ pư} = m_{CaSO_3} + m_{dd\ HCl} - m_{SO_2} = 12 + 50 - 6,4 = 55,6(gam)$
$C\%_{CaCl_2} = \dfrac{0,1.111}{55,6}.100\% = 19,96\%$
Ta có: \(n_{CaSO_3}=\dfrac{12}{120}=0,1\left(mol\right)\)
a. PTHH: CaSO3 + 2HCl ---> CaCl2 + H2O + SO2
b. Theo PT: \(n_{SO_2}=n_{CaSO_3}=0,1\left(mol\right)\)
=> \(m_{SO_2}=0,1.64=6,4\left(g\right)\)
c. Theo PT: \(n_{HCl}=2.n_{CaSO_3}=2.0,1=0,2\left(mol\right)\)
=> \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
Ta có: \(C_{\%_{HCl}}=\dfrac{7,3}{m_{dd_{HCl}}}.100\%=14,6\%\)
=> \(m_{dd_{HCl}}=50\left(g\right)\)
d. Ta có: \(m_{dd_{CaCl_2}}=12+50-0,1.64=55,6\left(g\right)\)
Theo PT: \(n_{CaCl_2}=n_{SO_2}=0,1\left(mol\right)\)
=> \(m_{CaCl_2}=0,1.111=11,1\left(g\right)\)
=> \(C_{\%_{CaCl_2}}=\dfrac{11,1}{55,6}.100\%=19,96\%\)
a. PTHH: Zn + H2SO4 ---> ZnSO4 + H2↑
b. Ta có: \(C_{M_{H_2SO_4}}=\dfrac{n_{H_2SO_4}}{200:1000}=1,5M\)
=> \(n_{H_2SO_4}=0,3\left(mol\right)\)
Ta lại có: \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)
Ta thấy: \(\dfrac{0,3}{1}>\dfrac{0,25}{1}\)
Vậy H2SO4 dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)
=> \(V_{H_2}=0,25.22,4=5,6\left(lít\right)\)
c. Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,25\left(mol\right)\)
=> \(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)
d. Ta có: \(V_{dd_{ZnSO_4}}=0,2\left(lít\right)\)
=> \(C_{M_{ZnSO_4}}=\dfrac{0,25}{0,2}=1,25M\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
1 1 1 1
0,3 0,3 0,3 0,3
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
a). \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒\(V_{H2}=n.22,4=0,3.22,4=6,72\left(l\right)\)
b). \(80ml=0,08l\)
\(n_{H2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
→\(C_M=\dfrac{n}{V}=\dfrac{0,3}{0,08}=3,75\left(M\right)\)
c). \(n_{MgSO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{MgSO4}=n.22,4=0,3.22,4=6,72\left(l\right)\)
→\(C_M=\dfrac{n}{V}=\dfrac{0,3}{6,72}=0,04\left(M\right)\)
d). \(MgSO_4+Ba\left(OH\right)_2\rightarrow Mg\left(OH\right)_2+BaSO_4\downarrow\)
1 1 1 1
0,3 0,3 0,3
\(n_{BaSO4\uparrow}=\dfrac{0,3.1}{1}\)=0,3(mol)
→\(m_{BaSO4\downarrow}=n.M=0,3.233=69,9\left(g\right)\)
\(n_{Ba\left(OH\right)_2}=\dfrac{0,3.1}{1}\)=0,3(mol)
\(\rightarrow V_{ddBa\left(OH\right)_2}=\dfrac{n}{C_M}=\dfrac{0,3}{1,6}=0,1875\left(l\right)\)
a)
$Mg + H_2SO_4 \to MgSO_4 + H-2$
b) $n_{H_2SO_4} = n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$C\%_{H_2SO_4} = \dfrac{0,2.98}{200}.100\% = 9,8\%$
$n_{H_2} = n_{Mg} = 0,2(mol)$
$\Rightarrow m_{dd\ A} = 4,8 + 200 - 0,2.2 = 204,4(gam)$
$C\%_{MgSO_4} = \dfrac{0,2.120}{204,4}.100\% = 11,7\%$
c) $V_{H_2} = 0,2.22,4 = 4,48(lít)$
\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,V_{H_2\left(25\text{đ}\text{ộ}C,1bar\right)}=0,2.24,79=4,958\left(l\right)\\ c,n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{\text{dd}HCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ m_{\text{dd}A}=m_{Mg}+m_{\text{dd}HCl}-m_{H_2}=4,8+146-0,2.2=150,4\left(g\right)\\ d,C\%_{\text{dd}MgCl_2}=\dfrac{95.0,2}{150,4}.100\approx12,633\%\)