\(Mg+2HCl \rightarrow MgCl_2+H_2\\ MgCl_2+2NaOH \rightarrow Mg(OH)_2+2NaCl\\ n_{HCl}=0,1mol\\ n_{Mg}=0,05mol\\ n_{MgCl_2}=n_{Mg}=0,05mol\\ n_{NaOH}=0,06mol\\ MgCl_2: 0,05>NaOH:\frac{0,06}{2}=0,03 \Rightarrow \text{MgCl2 dư, NaOH hết}\\ n_{Mg(OH)_2}=\frac{1}{2}NaOH=\frac{1}{2}.0,06=0,03mol\\ m_{Mg(OH)_2}=0,03.58=1,74g \)

\(\text{Em sửa 7,1 thành 8,1g nha}\\ a/ 2Al+3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2\\ b/ \\ n_{Al}=0,3mol\\ n_{H_2SO_4}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,3=0,45mol V_{H_2SO_4}=\frac{0,45}{0,05}=9M\\ c/ \\ V_{H_2}=0,45.22,4=10,08l\)

\(n_{H_2}=0,045mol\\ \text{Bảo toàn nguyên tố H:}\\ 2.n_{H_2}=n_{HCl}=0,09mol\\ \text{BTKL}\\ m_{hh}+m_{HCl}=m_{Khan}+m_{H_2}\\ m_{hh}+0,09.36,5=4,575+0,045.2\\ m_{hh}=1,38g\)

\(Mg+2HCl \rightarrow MgCl_2+H_2\\ Fe+2HCl \rightarrow FeCl_2+H_2\\ n_a=Mg\\ n_b=Fe\\ n_{HCl}=\frac{50.29,2\%}{36,5}=0,4mol\\ m_{hh}=24a+56b=19,2(1)\\ n_{HCl}=2a+2b=0,4(2)\\ (1)(2)\\ \)

Xem lại đè bài nha

\(2Al+6HCl \rightarrow 2AlCl_3+3H_2\\ Mg+2HCl \rightarrow MgCl_2+H_2\\ n_a=Al\\ n_b=Mg\\ m_{hh}=27a+24b=7,8(1)\\ m_{muối}=133,5a+95b=36,2(2)\\ (1)(2)\\ a=0,2\\ b=0,1\\ n_{H_2}=1,5a+b=1,5.0,2+0,1=0,4mol\\ V_{H_2}=0,4.22,4=8,96l\)

\(SO_3+H_2O\rightarrow H_2SO_4\\ Ba+H_2SO_4 \rightarrow BaSO_4+H_2\\ n_{SO_3}=\frac{8}{80}=0,1mol\\ n_{H_2O}=\frac{32}{18}=2mol\\ SO_3 < H_2O\\ n_{SO_3}=n_{H_2SO_4}=0,1mol\\ n_{Ba}=0,1mol\\ H_2SO_4=Ba\\ n_{BaSO_4}=0,1.233=23,3g\)

\(R+2HCl \rightarrow RCl_2+H_2\\ n_{H_2}=\frac{2,24}{22,4}=0,1mol\\ M_R=\frac{5,6}{0,1}=56 g/mol\\ \Rightarrow R: Fe\)

Câu B

A sai vì KCl là muối

C sai vì O2 là khí

D xài vì NO2 là khí

\(2Na+2HCl\rightarrow 2NaCl+H_2\\ n_{Na}=\frac{5,75}{23}=0,25mol\\ n_{Na}=n_{HCl}=0,25mol\\ m_{HCl_{dd}}=\frac{0,25.36,5.100}{18,25}=50g\)

\(n_{Al}=\frac{8,1}{27}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2(SO_4)_3 +3H_2 n_{H_2}=0,45mol\\ V=10,08l\)