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\(n_{NaCl}=\dfrac{3,51}{58,5}=0,06\left(mol\right)\)
a) Pt : \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl|\)
1 1 1 1
0,06 0,06 0,06 0,06
a) \(n_{AgCl}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)
⇒ \(m_{AgCl}=0,06.143,5=8,61\left(g\right)\)
b) \(n_{AgNO3}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)
\(V_{ddAgNO3}=\dfrac{0,06}{0,2}=0,3\left(l\right)\)
c) \(n_{NaNO3}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)
\(C_{M_{NaNO3}}=\dfrac{0,06}{0,3}=0,2\left(M\right)\)
Chúc bạn học tốt
\(a,n_{NaCl}=\dfrac{3,51}{58,5}=0,06(mol)\\ PTHH:NaCl+AgNO_3\to AgCl\downarrow+NaNO_3\\ \Rightarrow n_{AgCl}=0,06(mol)\\ \Rightarrow m_{AgCl}=0,06.143,5=8,61(g)\\ b,n_{AgNO_3}=0,06(mol)\\ \Rightarrow V_{dd_{AgNO_3}}=\dfrac{0,06}{0,2}=0,3(l)\\ c,n_{NaNO_3}=0,06(mol);V_{dd_{NaNO_3}}=V_{dd(\text {phản ứng})}=0,3(l)\\ \Rightarrow C_{M_{NaNO_3}}=\dfrac{0,06}{0,3}=0,2M\)
\(m_{H_2SO_4}=\dfrac{19,6\cdot20\%}{100\%}=3,92\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\\ PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \Rightarrow n_{H_2SO_4}=n_{BaCl_2}=n_{BaSO_4}=0,04\left(mol\right)\\ \Rightarrow m_{CT_{BaCl_2}}=0,04\cdot208=8,32\left(g\right)\\ \Rightarrow m_{dd_{BaCl_2}}=\dfrac{8,32\cdot100\%}{12\%}\approx69,3\left(g\right)\\ m_{kết.tủa}=m_{BaSO_4}=0,04\cdot233=9,32\left(g\right)\)
Bài 10:
PTHH: \(Na_2CO_3+BaCl_2\rightarrow2NaCl+BaCO_3\downarrow\)
a) Ta có: \(n_{Na_2CO_3}=\dfrac{200\cdot10,6\%}{106}=0,2\left(mol\right)=n_{BaCO_3}\)
\(\Rightarrow m_{BaCO_3}=0,2\cdot197=39,4\left(g\right)\)
b) Theo PTHH: \(n_{BaCl_2}=n_{BaCO_3}=0,2mol\)
\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,2\cdot208}{120}\cdot100\%\approx34,67\%\)
c) Theo PTHH: \(n_{NaCl}=2n_{BaCl_2}=0,4mol\) \(\Rightarrow m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)
Mặt khác: \(m_{dd}=m_{ddNa_2CO_3}+m_{ddBaCl_2}-m_{BaCO_3}=280,6\left(g\right)\)
\(\Rightarrow C\%_{NaCl}=\dfrac{23,4}{280,6}\cdot100\%\approx8,34\%\)
Bài 14 :
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,15 0,3 0,15 0,15
a) \(n_{H2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)
b) \(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)
c) \(n_{FeCl2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
\(C_{M_{FeCl2}}=\dfrac{0,15}{0,15}=1\left(M\right)\)
Chúc bạn học tốt
ở đoạn c bạn có ghi nhầm ko à , tại mình cứ thấy nó sai sai
a) \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
b) \(n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\)
PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
0,25----->0,25------->0,25---->0,5
=> \(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
=> \(m_{ddH_2SO_4}=\dfrac{24,5.100}{19,6}=125\left(g\right)\)
c) \(m_{BaSO_4}=0,25.233=58,25\left(g\right)\)
d)
\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Xét tỉ lệ \(\dfrac{0,5}{1}>\dfrac{0,2}{1}\) => NaOH hết, HCl dư
=> Quỳ tím chuyển màu đỏ
\(a)n_{BaCl_2}=\dfrac{240}{1,12}:1000\cdot1=\dfrac{3}{14}mol\\ n_{H_2SO_4}=\dfrac{122.20}{100}:98=\dfrac{61}{245}mol\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ \Rightarrow\dfrac{3:14}{1}< \dfrac{61:245}{1}\Rightarrow H_2SO_4.dư\\ n_{BaSO_4}=n_{BaCl_2}=n_{H_2SO_4}=\dfrac{3}{14}mol\\ m_{kt}=m_{BaSO_4}=\dfrac{3}{14}\cdot233=50g\\ c)C_{\%H_2SO_4\left(dư\right)}=\dfrac{\left(61:245-3:14\right)98}{240+122-50}\cdot100=1,2\%\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
___0,1______________0,1____0,1 (mol)
\(FeSO_4+BaCl_2\rightarrow FeCl_2+BaSO_{4\downarrow}\)
__0,1______0,1_____________0,1 (mol)
a, VH2 = 0,1.22,4 = 2,24 (l)
b, \(V_{BaCl_2}=\dfrac{0,1}{0,2}=0,5\left(l\right)\)
c, \(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
Bạn tham khảo nhé!