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a, \(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
\(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)
Lần sau bạn đăng tách từng bài ra nhé.
Câu 1:
a, \(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
Câu 3: \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
\(n_{CaCO_3}=n_{CO_2}=0,4\left(mol\right)\Rightarrow m_{CaCO_3}=0,4.100=40\left(g\right)\)
Câu 4: \(n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)
PT: \(CuSO_4+BaCl_2\rightarrow BaSO_{4\downarrow}+CuCl_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\), ta được CuSO4 dư.
Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
a)
$Na_2O + H_2O \to 2NaOH$
n Na2O = 15,5/62 = 0,25(mol)
n NaOH = 2n Na2O = 0,5(mol)
=> CM NaOH = 0,5/0,5 = 1M
b) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
n H2SO4 = 1/2 n NaOH = 0,25(mol)
=> m dd H2SO4 = 0,25.98/20% = 122,5(gam)
=> V dd H2SO4 = m / D = 122,5/1,14 =107,46(ml)
c) n Na2SO4 = n H2SO4 = 0,25(mol)
CM Na2SO4 = 0,25/0,10746 = 2,33M
a)
`\(Na_2O++H_{2_{ }}O->2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25mol\)
\(n_{Na_2O}=2n_{Na_2O}=0,5mol\)
\(C_{M_{NaOH}}=\dfrac{0,5}{0,5}\)=1M
nNa2O = 0,125 mol
a. Na2O + H2O --------> NaOH
0,125 mol ----------------> 0,125 mol
--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M
b. H2SO4 + 2NaOH ------> Na2SO4 + H2O
....0,0625 <---0,125 mol
--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g
--> mH2SO4(20%) = 6,125/20% = 30,625 g
suy ra V = m/D = 30,625 / 1,14 = 26,86 ml
nNa2O = 0,125 mol
a. Na2O + H2O --------> NaOH
0,125 mol ----------------> 0,125 mol
--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M
b. H2SO4 + 2NaOH ------> Na2SO4 + H2O
....0,0625 <---0,125 mol
--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g
--> mH2SO4(20%) = 6,125/20% = 30,625 g
suy ra V = m/D = 30,625 / 1,14 = 26,86 ml
1. \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
PTHH: Na2O + H2O → 2NaOH
Mol: 0,25 0,5
\(C_{M_{ddNaOH}}=\dfrac{0,5}{0,5}=1M\)
2.
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,5 0,25
\(m_{ddH_2SO_4}=\dfrac{0,25.98.100}{20}=122,5\left(g\right)\)
\(V_{ddH_2SO_4}=\dfrac{122,5}{1,14}=107,456\left(ml\right)\)
\(n_{Na2O}=\dfrac{m_{Na2O}}{M_{Na2O}}=0,25\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,25 mol - 0,25 mol - 0,5 mol
a) \(C_{M_{NaOH}}=\dfrac{n_{NaOH}}{V_{NaOH}}=1\left(M\right)\)
b) \(H_2SO_4+2NaOH\rightarrow Na_2SO4+2H_2O\)
0,25 mol - 0,5 mol - 0,25 mol - 0,5 mol
\(m_{ctH2SO4}=n_{H2SO4}.M_{H2SO4}=24,5\left(g\right)\)
\(C_{\%_{H2SO4}}=\dfrac{m_{ctH2SO4}}{m_{ddH2SO4}}.100\%\)
\(\Rightarrow m_{ddH2SO4}=\dfrac{m_{ctH2SO4}.100\%}{C_{\%_{H2SO4}}}=122,5\left(g\right)\)
\(D_{H2SO4}=\dfrac{m_{ddH2SO4}}{V_{H2SO4}}\Rightarrow V_{H2SO4}=\dfrac{m_{ddH2SO4}}{D_{H2SO4}}\approx107,46\left(ml\right)\)
a)
\(n_{Na_2O}=\dfrac{31}{62}=0,5\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
______0,5--------------->1
=> \(C_{M\left(NaOH\right)}=\dfrac{1}{0,5}=2M\)
b)
PTHH: H2SO4 + 2NaOH --> Na2SO4 + 2H2O
______0,5<---------1
=> mH2SO4 = 0,5.98 = 49(g)
=> \(m_{dd\left(H_2SO_4\right)}=\dfrac{49.100}{20}=245\left(g\right)\)
=> \(V_{dd\left(H_2SO_4\right)}=\dfrac{245}{1,14}=214,912\left(ml\right)\)
\(n_{Na_2O}=\dfrac{31}{62}=0,5(mol)\\ a,Na_2O+H_2O\to 2NaOH\\ \Rightarrow n_{NaOH}=1(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1}{0,5}=2M\\ b,2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=0,5(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,5.98}{20\%}=245(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{245}{1,14}=214,91(ml)\)
BaO+H2O -> Ba(OH)2
0,02 0,02
a) CM = n/V = 0,02/0,02 = 1M
b) Ba(OH)2 + H2SO4 -> BaSO4 +2H2O
0,02 0,02
=> m = 0,392 g
D = m/V = 1,14
=> 0,392/V = 1,14 => V = 0,34l
a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)