Question

Cho 200gam dd NaOH 6% vào 300ml dd H2SO4 0,25M. Tìm khối lượng muối thu được ( Na2SO4)

Answer 1

\(n_{NaOH}=\dfrac{6\%.200}{40}=0,3\left(mol\right)\\ n_{H_2SO_4}=0,25.0,3=0,075\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O

LTL: \(\dfrac{0,3}{2}>0,075\) => NaOH dư

\(n_{Na_2SO_4}=0,075\left(mol\right)\\ m_{Na_2SO_4}=0,075.142=10,65\left(g\right)\)

Answer 2

Answer 3

\(n_{NaOH}=\dfrac{200.6\%}{40}=0,3\left(mol\right)\\ n_{H_2SO_4}=0,3.0,25=0,075\left(mol\right)\\ pthh:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\) 
 \(LTL:\dfrac{0,3}{2}>\dfrac{0,075}{1}\) 

=> NaOH dư , H2SO4 hết 
theo pthh : \(n_{Na_2SO_4}=\dfrac{1}{2}n_{H_2SO_4}=0,0375\left(mol\right)\) 
=> \(m_{Na_2SO_4}=0,0375.142=5,325\left(g\right)\)