40 ml = 0,04 lít
60 ml = 0,06 lít
a) Na2SO4 + BaCl2 \(\rightarrow\) BaSO4 + 2NaCl
b) Ta có: \(C_M=\dfrac{n_{Na_2SO_4}}{V_{Na_2SO_4}}=\dfrac{n}{0,04}=0,5\)
\(\Rightarrow n_{Na_2SO_4}=0,02\left(mol\right)\)
Ta có: \(C_M=\dfrac{n_{BaCl_2}}{V_{BaCl_2}}=\dfrac{n}{0,06}=0,5\)
\(\Rightarrow n_{BaCl_2}=0,03\left(mol\right)\)
Ta thấy: \(\dfrac{0,02}{1}< \dfrac{0,03}{1}\)
=> Na2SO4 hết, BaCl2 dư
\(\Rightarrow n_{BaSO_4}=0,02\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,02.233=4,66\left(g\right)\)
c) \(\Rightarrow n_{BaCl_2}dư=0,03-0,02=0,01\left(mol\right)\)
Vdd sau phản ứng là: 0,04 + 0,06 = 0,1 (lít)
=> \(C_M=\dfrac{n_{BaSO_4}}{V_{spu}}=\dfrac{0,02}{0,1}=0,2M\)
\(C_M=\dfrac{n_{BaCl_2}}{V_{spu}}=\dfrac{0,01}{0,1}=0,1M\)
