Bài 1:
\(n_{NaOH}=\dfrac{16}{40}=0,4(mol)\\ PTHH:NaOH+HCl\to NaCl+H_2O\\ \Rightarrow n_{NaCl}=n_{NaOH}=0,4(mol)\\ \Rightarrow m_{NaCl}=0,4.58,5=23,4(g)\)
Bài 2:
\(n_{Ca(OH)_2}=1.0,5=0,5(mol)\\ PTHH:Ca(OH)_2+2HCl\to CaCl_2+2H_2O\\ \Rightarrow n_{HCl}=2n_{Ca(OH)_2}=1(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{1.36,5}{200}.100\%=18,25\%\\ \Rightarrow a=18,25\)
Bài 3:
\(n_{H_2SO_4}=\dfrac{50.9,8\%}{100\%.98}=0,05(mol)\\ PTHH:BaCl_2+H_2SO_4\to BaSO_4\downarrow+2HCl\\ \Rightarrow n_{BaSO_4}=n_{H_2SO_4}=0,05(mol)\\ \Rightarrow a=m_{BaSO_4}=0,05.233=11,65(g)\)
Bài 4:
\(n_{CaCO_3}=\dfrac{20}{100}=0,2(mol)\\ PTHH:CaCO_3+2HCl\to CaCl_2+H_2O+CO_2\uparrow\\ \Rightarrow n_{CO_2}=n_{CaCO_3}=0,2(mol)\\ \Rightarrow V_{CO_2}=0,2.22,4=4,48(l)\)
