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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: \(n_{Na_2SO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)
=> \(m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
=> \(C_{\%_{Na_2SO_4}}=\dfrac{28,4}{240}.100\%=11,83\%\)
\(n_{Na_2CO_3}=0,1.1=0,1\left(mol\right)\)
a. \(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)
0,1 0,1 0,1 0,2
b. \(m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\)
c. \(C\%_{Ba\left(OH\right)_2}=\dfrac{0,1.171.100}{200}=8,55\%\)
d. \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\)
0,1 0,2
=> \(a=m_{dd.HCl}=\dfrac{0,2.36,5.100}{30}=\dfrac{73}{3}\left(g\right)\)
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.
\(n_{Ba\left(OH\right)_2}=n_{Na_2CO_3}=0,2\left(mol\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{0,2.171.100\%}{200}=17,1\%\)
\(n_{BaCl_2}=\dfrac{150.16,64\%}{137+35.2}=0,12\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.14,7\%}{98}=0,15\left(mol\right)\)
Phương trình hóa học :
BaCl2 + H2SO4 -----> BaSO4 + 2HCl
Dễ thấy \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{1}\Rightarrow H_2SO_4\text{ dư }0,15-0,12=0,03\left(mol\right)\)
c) Khối lượng kết tủa :
\(m_{BaSO_4}=0,12.233=27,96\) (g)
Khối lượng chất tan : \(m_{HCl}=0,24.36,5=8,76\left(g\right)\) ;
\(m_{H_2SO_4\left(\text{dư}\right)}=0,03.98=2,94\left(g\right)\)
c) \(C\%_{H_2SO_4}\)= \(\dfrac{2,94}{150+100}.100\%=1,176\%\)
\(C\%_{HCl}=\dfrac{8,76}{150+100}.100\%=3.504\%\)
d) NaOH + HCl ---> NaCl + H2O
0,24 <-- 0,24
mol mol
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
0,06 mol <-- 0,03 mol
\(\Rightarrow n_{NaOH}=0,24+0,06=0,3\left(mol\right)\)
\(V_{NaOH}=0,3.2=0,6\left(l\right)\)
Bài 1:
PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Ta có: \(\left\{{}\begin{matrix}m_{Ba\left(OH\right)_2}=150\cdot17,1\%=25,65\left(g\right)\\m_{HCl}=300\cdot7,3\%=21,9\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=\frac{25,65}{171}=0,15\left(mol\right)\\n_{HCl}=\frac{21,9}{36,5}=0,6\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\frac{0,15}{1}< \frac{0,6}{2}\) \(\Rightarrow\) Ba(OH)2 phản ứng hết, HCl còn dư
\(\Rightarrow\) Dung dịch A làm quỳ tím hóa đỏ
Bài 3:
PTHH: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_4\downarrow\) (1)
a) Ta có: \(\left\{{}\begin{matrix}n_{BaCl_2}=\frac{150\cdot5,2\%}{208}=0,0375\left(mol\right)\\n_{H_2SO_4}=\frac{250\cdot19,6\%}{98}=0,5\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\frac{0,0375}{1}< \frac{0,5}{1}\) \(\Rightarrow\) BaCl2 phản ứng hết, H2SO4 còn dư
\(\Rightarrow n_{BaSO_4}=0,0375mol\) \(\Rightarrow m_{BaSO_4}=0,0375\cdot233=8,7375\left(g\right)\)
b) Dung dịch A chứa \(HCl\) và \(H_2SO_{4\left(dư\right)}\)
PTHH: \(HCl+NaOH\rightarrow NaCl+H_2O\) (2)
\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\) (3)
Theo PTHH (1): \(\left\{{}\begin{matrix}n_{HCl}=2n_{BaCl_2}=0,075mol\\n_{H_2SO_4\left(dư\right)}=0,4625mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH\left(2\right)}=0,075mol\\n_{NaOH\left(3\right)}=0,925mol\end{matrix}\right.\)
\(\Rightarrow n_{NaOH}=1mol\) \(\Rightarrow V_{NaOH}=\frac{1}{1,5}\approx0,67\left(l\right)=670\left(ml\right)\)
Mik làm rồi nhé
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