a,b)\(n_{H_2SO_4}=0,2\) theo PT \(2n_{H_2SO_4}=n_{NaOV}=0,4\Rightarrow m_{NaOV}=0,4.4016kg\)
\(H_2SO_4+2NaOH\rightarrow H_2Na_2SO_4+H_2\)
\(m_{ddNaOH}=\dfrac{16}{2}=80g\)
c)\(n_{KOH}=2_{n_{H_2SO_4}}=0,4\Rightarrow m_{KOH}=0,3.56-22,4g\)
\(\Rightarrow m_{KOH\left(dd\right)}=\dfrac{22,4}{0,056}=400\Rightarrow V=\dfrac{400}{1,045}=182,8\left(ml\right)\)
a, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\) \(\left(1\right)\)
b, Đổi \(20ml=0,02l\)
\(n_{H_2SO_4}=0,02.1=0,02\left(mol\right)\)
Thep phương trình \(\left(1\right)\) ta được:
\(n_{NaOH}=2n_{H_2SO_4}=2.0,02=0,04\left(mol\right)\\ \Rightarrow m_{NaOH}=0,04.40=1,6\left(g\right)\\ \Rightarrow m_{dd NaOH}=\dfrac{1,6}{20}.100=8\left(g\right)\)
c, \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\) \(\left(2\right)\)
Theo phương trình \(\left(2\right)\):
\(n_{KOH}=2n_{H_2SO_4}=2.0,02=0,04\left(mol\right)\\ \Rightarrow m_{KOH}=0,04.56=2,24\left(g\right)\\ \Rightarrow m_{dd KOH}=\dfrac{2,24}{5,6}.100=40\left(g\right)\\ \Rightarrow V_{dd KOH}=\dfrac{40}{1.045}=38,3\left(ml\right)\)
