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nBaCl2 1M = 0.5 (mol)
nBaCl2 2.5M = 1.25 (mol)
nBaCl2 sau khi trộn = 1.75 (mol)
Vdd sau khi trộn = 500ml + 500ml = 1 (l)
CMdd sau khi trộn = n/V = 1.75/1 = 1.75M
50 gam dung dich BaCl2 10% có:
\(m_{BaCl_2}=\dfrac{10.50}{100}=5\left(g\right)\)
70 gam dung dịch BaCl2 20% có:
\(m_{BaCl_2}=\dfrac{20.70}{100}=14\left(g\right)\)
Sau khi trộn thì:
\(m_{ddsau}=50+70=120\left(g\right)\)
\(m_{BaCl_2\left(sau\right)}=5+14=19\left(g\right)\)
\(\Rightarrow C\%_{BaCl_2\left(sau\right)}=\dfrac{19}{120}.100\%=15,83\%\)
nHCl(1)=0,5.0,2=0,1 mol
nHCl(2)=0,2.0,3=0,06 mol
VddHCl sau khi trộn=500+200=700ml=0,7 lít
Tổng nHCl sau khi trộn=0,1+0,06=0,16 mol
CM dd HCl sau khi trộn=0,16/0,7=0,23M
H2SO4 + BaCl2 → BaSO4↓ + 2HCl
\(n_{H_2SO_4}=0,5\times2=1\left(mol\right)\)
Theo PT: \(n_{BaSO_4}=n_{H_2SO_4}=1\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=1\times233=233\left(g\right)\)
Theo PT: \(n_{HCl}=2n_{H_2SO_4}=2\times1=2\left(mol\right)\)
\(\Rightarrow m_{HCl}=2\times36,5=73\left(g\right)\)
\(n_{Na_2CO_3}=\dfrac{200.13,25%}{100\%.106}=0,25(mol)\\ n_{BaCl_2}=\dfrac{20,8\%.300}{100\%.208}=0,3(mol)\\ PTHH:Na_2CO_3+BaCl_2\to BaSO_4\downarrow +2NaCl\)
Vì \(\dfrac{n_{Na_2CO_3}}{1}<\dfrac{n_{BaCl_2}}{1}\) nên \(BaCl_2\) dư
Do đó dd sau p/ứ gồm \(BaCl_2\) dư và \(NaCl\)
\(n_{BaSO_4}=0,25(mol);n_{NaCl}=0,5(mol);n_{BaCl_2(dư)}=0,3-0,25=0,05(mol)\\ \Rightarrow m_{NaCl}=0,5.58,5=29,25(g);m_{BaCl_2(dư)}=0,05.208=10,4(g)\\ \Rightarrow C\%_{NaCl}=\dfrac{29,25}{200+300-0,25.233}.100\%=6,62\%\\ C\%_{BaCl_2(dư)}=\dfrac{10,4}{200+300-0,25.233}.100\%=2,35\%\)
\(n_{BaCl_2}=\dfrac{41,6}{208}=0,2\left(mol\right)\)
\(n_{AgNO_3}=\dfrac{17}{170}=0,1\left(mol\right)\)
\(BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+2AgCl\)
\(\dfrac{0,2}{1}>\dfrac{0,1}{2}\) ⇒ BaCl2 dư.
a, \(n_{AgCl}=n_{AgNO_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,1.143,5=14,35\left(g\right)\)
b, \(n_{Ba\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgNO_3}=0,05\left(mol\right)\)
\(C_{M_{Ba\left(NO_3\right)_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(n_{BaCl_2phan/ung}=\dfrac{1}{2}n_{AgNO_3}=0,05\left(mol\right)\)
\(\Rightarrow n_{BaCl_2dư}=0,15\left(mol\right)\rightarrow C_{M\left(BaCl_2\right)}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)
nBaCl2 1M = 0.5 x 1 = 0.5 (mol)
nBaCl2 2.5M = 2.5 x 0.5 = 1.25 (mol)
nBaCl2 sau khi trộn: 0.5 + 1.25 = 1.75 (mol)
Vdd sau khi trộn: 500ml + 500ml = 1000ml = 1l
CMdd sau khi trộn: 1.75/1 = 1.75 M
nBaCl2 (1)= 0.5 mol
nBaCl2 (2)= 1.25mol
nBaCl2 (1,2)= 1.75mol
VBaCl2 (1,2)= 0.5+0.5=1l
CM BaCl2= 1.75/1=1.75M