\(\to C\\ CuO+H_2 \to Cu+H_2O\)

\(1/ K_2CO_3\\ PTK: 39.2+12+16.3=138 (đvc)\\ 2/ Al(NO_3)_3\\ PTK: 27+(14+16.3).3=213(đvc)\\ 3/ FeSO_4\\ PTK: 56+32+16.4=152(đvc)\\ 4/ R_2O_n\\ PTK: R.2+16.n (đvc)\)

\(Mg+2HCl \to MgCl_2+H_2\\ n_{H_2}=\frac{11,2}{22,4}=0,5(mol)\\ a/\\ n_{Mg}=n_{H_2}=0,5(mol)\\ m_{Mg}=0,5.24=12(g)\\ b/\\ n_{HCl}=2.n_{H_2}=2.0,5=1(mol) V_{HCl}=\frac{1}{2}=0,5(l)\)

Xem xong bt đt dạng sơ đồ mà @@

\(4P+5O_2\buildrel{{t^o}}\over\longrightarrow 2P_2O_5\\ n_{P_2O_5}=\frac{28,4}{142}=0,2(mol)\\ n_P=2n_{P_2O_5}=0,2.2=0,4(mol)\\ a/ m_P=0,4.31=12,4(g)\\ b/\\ n_{O_2}=2,5.n_{P_2O_5}=2,5.0,2=0,5(mol)\\ V_{O_2}=0,5.22,4=11,2(l)\\ V_{kk}=5.V_{O_2}=11,2.5=56(l) \)

lâu lâu lộn hih :>

à lộn nhầm :v

\(n_{CO_2}=0,12mol\\ n_{NaOH}=0,2mol\\ T=\frac{n_{NaOH}}{n_{CO_2}}=\frac{0,2}{0,12}=1,66// \to Na2CO_3; NaHCO_3// 2NaOH+CO_2 \to Na_2CO_3\\ NaoH+CO_2 \to NaHCO_3\\ n_{Na_2CO_3}=a(mol)\\ n_{NaHCO_3}=b(mol)\\ n_{NaOH}=2a+b=0,2(1)\\ n_{CO_2}=a+b=0,12(2)\\ (1)(2)\\ a=0,08; b=0,04\\ CM_{Na_2CO_3}=\frac{0,08}{0,2}=0,4(mol)\\ CM_{NaHCO_3}=\frac{0,04}{0,2}=0,2(mol)\\ \)

Có phương pháp đây à :)

ko làm đc @@ nó cho 4 mol cua ion hết r bt đt sao ra