Trong mỗi phần có khối lượng là $\frac{12,1}{2}$= 6,05(g) $4Al + 3O_{2} \rightarrow 2Al_{2}O_{3}$(mol) a : 0,75a $2Mg + O_{2} \rightarrow 2MgO$(mol) b : 0,5b $4Na + O_{2} \rightarrow 2Na_{2}O$(mol) c : 0,25cTheo định luật bảo toàn khối lượng: $m_{Kim loại} + m_{O_{2}} = m_{Oxit}$$\Rightarrow 6,05 + m_{O_{2}}=9,65$ $\Rightarrow m_{O_{2}}=3,6 (g) \Rightarrow n_{O_{2}}=0,1125 (mol)$$\Rightarrow 0,75a+0,5b+0,25c=0,1125$ $2Al+6HCl \rightarrow 2AlCl_{3}+3H_{2}$(mol)a : 3a : :1,5a $Mg+2HCl \rightarrow MgCl_{2}+H_{2}$(mol)b : 2b : : b $2Na+2HCl \rightarrow 2NaCl + H_{2}$(mol)c : c : : 0,5aTa có:$n_{H_{2}}=1,5a+b+0,5c=2.(0,75a+0,5b+0,25c)=2.0,1125=0,225(mol)$$\Rightarrow V=0,225.22,4=5,04(l)$$n_{Cl}=n_{HCl}=2n_{H_{2}}=0,225.2=0,45(mol)$$\Rightarrow m = m_{kl}+m_{Cl}=6,05+0,45.35,5=22,025(g)$
Bình chọn giảmTrong mỗi phần có khối lượng là 12,12" role="presentation" style="font-size: 12px; display: inline; position: relative;">12,1212,12= 6,05(g) 4Al+3O2→2Al2O3" role="presentation" style="font-size: 12px; display: inline; position: relative;">4Al+3O2→2Al2O34Al+3O2→2Al2O3(mol) a : 0,75a 2Mg+O2→2MgO" role="presentation" style="font-size: 12px; display: inline; position: relative;">2Mg+O2→2MgO2Mg+O2→2MgO(mol) b : 0,5b 4Na+O2→2Na2O" role="presentation" style="font-size: 12px; display: inline; position: relative;">4Na+O2→2Na2O4Na+O2→2Na2O(mol) c : 0,25cTheo định luật bảo toàn khối lượng: mKimloại+mO2=mOxit" role="presentation" style="font-size: 12px; display: inline; position: relative;">mKimloại+mO2=mOxitmKimloại+mO2=mOxit⇒6,05+mO2=9,65" role="presentation" style="font-size: 12px; display: inline; position: relative;">⇒6,05+mO2=9,65⇒6,05+mO2=9,65 ⇒mO2=3,6(g)⇒nO2=0,1125(mol)" role="presentation" style="font-size: 12px; display: inline; position: relative;">⇒mO2=3,6(g)⇒nO2=0,1125(mol)⇒mO2=3,6(g)⇒nO2=0,1125(mol)⇒0,7a+0,5b+0,25c=0,1125" role="presentation" style="font-size: 12px; display: inline; position: relative;">⇒0,75a+0,5b+0,25c=0,1125⇒0,7a+0,5b+0,25c=0,1125 2Al+6HCl→2AlCl3+3H2" role="presentation" style="font-size: 12px; display: inline; position: relative;">2Al+6HCl→2AlCl3+3H22Al+6HCl→2AlCl3+3H2(mol)a : 3a : :1,5a Mg+2HCl→MgCl2+H2" role="presentation" style="font-size: 12px; display: inline; position: relative;">Mg+2HCl→MgCl2+H2Mg+2HCl→MgCl2+H2(mol)b : 2b : : b 2Na+2HCl→2NaCl+H2" role="presentation" style="font-size: 12px; display: inline; position: relative;">2Na+2HCl→2NaCl+H22Na+2HCl→2NaCl+H2(mol)c : c : : 0,5aTa có:nH2=1,5a+b+0,5c=2.(0,75a+0,5b+0,25c)=2.0,1125=0,225(mol)" role="presentation" style="font-size: 12px; display: inline; position: relative;">nH2=1,5a+b+0,5c=2.(0,75a+0,5b+0,25c)=2.0,1125=0,225(mol)nH2=1,5a+b+0,5c=2.(0,75a+0,5b+0,25c)=2.0,1125=0,225(mol)⇒V=0,225.22,4=5,04(l)" role="presentation" style="font-size: 12px; display: inline; position: relative;">⇒V=0,225.22,4=5,04(l)⇒V=0,225.22,4=5,04(l)nCl=nHCl=2nH2=0,225.2=0,45(mol)" role="presentation" style="font-size: 12px; display: inline; position: relative;">nCl=nHCl=2nH2=0,225.2=0,45(mol)nCl=nHCl=2nH2=0,225.2=0,45(mol)⇒m=mkl+mCl=6,05+0,45.35,5=22,025(g)" role="presentation" style="font-size: 12px; display: inline; position: relative;">⇒m=mkl+mCl=6,05+0,45.35,5=22,025(g)@shiro