Bài 4:
\(a,n_{KOH}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ \Rightarrow n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,05\left(mol\right)\\ \Rightarrow m_{K_2SO_4}=0,05\cdot174=8,7\left(g\right)\)
b, Sau phản ứng không thu được dung dịch
Bài 6:
\(a,PTHH:MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\\ Mg\left(OH\right)_2\rightarrow^{t^o}MgO+H_2O\\ b,\dfrac{n_{MgCl_2}}{1}>\dfrac{n_{NaOH}}{2}\Rightarrow MgCl_2.dư\\ n_{MgCl_2\left(dư\right)}=0,1-\dfrac{0,15}{2}=0,025\left(mol\right)\\ c,n_{MgO}=n_{Mg\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,075\left(mol\right)\\ \Rightarrow m=m_{MgO}=0,075\cdot40=3\left(g\right)\)