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\(n_{Na_2SO_4}=\dfrac{142.10}{100.142}=0,1(mol)\\ Na_2SO_4+Ba(OH)_2\to BaSO_4\downarrow+2NaOH\\ \Rightarrow n_{BaSO_4}=n_{Ba(OH)_2}=0,1(mol);n_{NaOH}=0,2(mol)\\ a,m_{BaSO_4}=0,1.233=23,3(g)\\ b,m_{dd_{Ba(OH)_2}}=\dfrac{0,1.171}{15\%}=114(g)\\ c,C\%_{NaOH}=\dfrac{0,2.40}{142+114-23,3}.100\%=3,44\%\)
Ta có: \(n_{Na_2SO_4}=\dfrac{\dfrac{10\%.142}{100\%}}{142}=0,1\left(mol\right)\)
\(PTHH:Na_2SO_4+Ba\left(OH\right)_2--->BaSO_4\downarrow+2NaOH\)
a. Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=n_{Na_2SO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
b. Ta có: \(m_{Ba\left(OH\right)_2}=0,1.171=17,1\left(g\right)\)
Mà: \(C_{\%_{Ba\left(OH\right)_2}}=\dfrac{17,1}{m_{dd_{Ba\left(OH\right)_2}}}.100\%=15\%\)
\(\Leftrightarrow m_{dd_{Ba\left(OH\right)_2}}=114\left(g\right)\)
c. Ta có: \(m_{dd_{NaOH}}=114+14,2-23,3=104,9\left(g\right)\)
Theo PT: \(n_{NaOH}=2.n_{Ba\left(OH\right)_2}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)
\(\Rightarrow C_{\%_{NaOH}}=\dfrac{8}{104,9}.100\%=7,63\%\)
\(a,n_{Na_2SO_4}=0,2\cdot0,2=0,04\left(mol\right);n_{Ba\left(OH\right)_2}=0,2\cdot0,1=0,02\left(mol\right)\\ PTHH:Na_2SO_4+Ba\left(OH\right)_2\rightarrow2NaOH+BaSO_4\downarrow\\ TL:....1.....1......2......1\left(mol\right)\\ BR:.......0,02.....0,02......0,04......0,02\left(mol\right)\)
Vì \(\dfrac{n_{Na_2SO_4}}{1}>\dfrac{n_{Ba\left(OH\right)_2}}{1}\) nên \(Na_2SO_4\) dư, \(Ba\left(OH\right)_2\) hết
\(b,C_{M_{NaOH}}=\dfrac{0,04}{0,2+0,2}=0,1M\)
a.
\(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)
b.
\(n_{BaCO_3}=n_{Na_2CO_3}=0,2.1=0,2\left(mol\right)\\ m_{kt}=197.0,2=39,4\left(g\right)\)
c.
\(n_{Ba\left(OH\right)_2}=n_{Na_2CO_3}=0,2\left(mol\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{0,2.171.100\%}{200}=17,1\%\)
câu 1
cho 2dd trên td vs NaOH dư
có tủa => CuSO4
CuSO4 + 2NaOH => Na2SO4 + Cu(OH)2
ko hiện tượng => Na2SO4
\(n_{Ba\left(OH\right)_2}=0,5.0,2=0,1\left(mol\right);n_{H_2SO_4}=0,3.0,4=0,12\left(mol\right)\)
PTHH: Ba(OH)2 + H2SO4 → BaSO4↓ + 2H2O
Mol: 0,1 0,1 0,1
Ta có: \(\dfrac{0,1}{1}< \dfrac{0,12}{1}\) ⇒ Ba(OH)2 hết, H2SO4 dư
\(C_{M_{H_2SO_4dư}}=\dfrac{0,12-0,1}{0,2+0,3}=0,04M\)
mdd sau pứ = 200.2,3+300.1,6-0,1.233 = 916,7 (g)
\(C\%_{H_2SO_4dư}=\dfrac{0,02.98.100\%}{916,7}=0,21\%\)
Ta có: \(C_{\%_{Ba\left(OH\right)_2}}=\dfrac{m_{Ba\left(OH\right)_2}}{250}.100\%=34,2\%\)
=> \(m_{Ba\left(OH\right)_2}=85,5\left(g\right)\)
=> \(n_{Ba\left(OH\right)_2}=\dfrac{85,5}{171}=0,5\left(mol\right)\)
Ta lại có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{150}.100\%=4,9\%\)
=> \(m_{H_2SO_4}=7,35\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{7,35}{98}=0,075\left(mol\right)\)
a. PTHH; Ba(OH)2 + H2SO4 ---> BaSO4↓ + 2H2O
Ta thấy: \(\dfrac{0,5}{1}>\dfrac{0,075}{1}\)
Vậy Ba(OH)2 dư.
Theo PT: \(n_{BaSO_4}=n_{H_2SO_4}=0,075\left(mol\right)\)
=> \(m_{BaSO_4}=0,075.233=17,475\left(g\right)\)
b. Ta có: \(m_{dd_{BaSO_4}}=250+7,35=257,35\left(g\right)\)
=> \(C_{\%_{BaSO_4}}=\dfrac{17,475}{257,35}.100\%=6,79\%\)
Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\)
a. PTHH: \(Ba\left(OH\right)_2+Na_2SO_4--->BaSO_4\downarrow+2NaOH\)
b. Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)
=> \(m_{BaSO_4}=0,2.233=46,6\left(g\right)\)
c. Theo PT: \(n_{Na_2SO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)
=> \(m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
=> \(C_{\%_{Na_2SO_4}}=\dfrac{28,4}{240}.100\%=11,83\%\)