$Ca+2H_2O\to Ca(OH)_2+H_2\uparrow$

$\begin{array}{l} 2Na+2H_2O\to 2NaOH+H_2\\ Ba+2H_2O\to Ba(OH)_2+H_2 \end{array}$

$\begin{array}{l} n_{P_2O_5}=\frac{28,4}{142}=0,2(mol)\\ P_2O_5+3H_2O\to 2H_3PO_4\\ n_{H_3PO_4}=2n_{P_2O_5}=0,4(mol)\\ n_{NaOH}=\frac{200.10\%}{40}=0,5(mol)\\ \to T=\frac{n_{NaOH}}{n_{H_3PO_4}}=\frac{0,5}{0,4}=1,25\\ \to Tao\,NaH_2PO_4:x(mol);Na_2HPO_4:y(mol)\\ NaOH+H_3PO_4\to NaH_2PO_4+H_2O\\ 2NaOH+H_3PO_4\to Na_2HPO_4+2H_2O\\ Theo\,PT:\,\begin{cases} x+y=0,4\\ x+2y=0,5 \end{cases}\to\begin{cases} x=0,3\\ y=0,1 \end{cases}\\ m_{dd\,spu}=200+28,4=228,4(g)\\ \to \begin{cases} C\%_{NaH_2PO_4}=\frac{0,3.120}{228,4}.100\%\approx 15,76\%\\ C\%_{Na_2HPO_4}=\frac{0,1.142}{228,4}.100\%\approx 6,22\% \end{cases} \end{array}$

$\begin{array}\\ n_{HNO_3}=\frac{6,3}{63}=0,1(mol)\\ N_2O_5+H_2O\to 2HNO_3\\ n_{N_2O_5}=n_{H_2O}=0,05(mol)\\ m=0,05.108=5,4(g);m_{H_2O}=0,05.18=0,9(g) \end{array}$

$\begin{array}\\ \%m_{Fe}=\frac{56}{152}.100\%\approx 36,8\%\\ \%m_S=\frac{32}{152}.100\%\approx 21,05\%\\ \%m_O=100-21,05-36,8=42,15\% \end{array}$

$m_{BaCl_2}=200.5\%=10(g)$

$\begin{array}{l} 850ml=0,85l\\ n_{KNO_3}=\frac{20}{101}\approx 0,2(mol)\\ C_{M\,KNO_3}=\frac{0,2}{0,85}\approx 0,235M \end{array}$

$\begin{array}{l} a)\\ n_{H_2}=\frac{4,48}{22,4}=0,2(mol)\\ 2K+2H_2O\to 2KOH+H_2\\ n_K=2n_{H_2}=0,4(mol)\\ \%m_K=\frac{0,4.39}{20,4}.100\%\approx 76,5\%\\ \%m_{Mg}=100-76,5=23,5\%\\ b)\\ n_{Mg}=\frac{20,4-0,4.39}{24}=0,2(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ 2K+2HCl\to 2KCl+H_2\\ \sum n_{H_2}=n_{Mg}+0,5n_K=0,4(mol)\\ V_{H_2}=0,4.22,4=8,96(l) \end{array}$

$\begin{array}{l} 100ml=0,1l\\ n_{NaOH}=0,1.1=0,1(mol)\\ C_6H_5OH+NaOH\to C_6H_5ONa+H_2O\\ n_{C_6H_5OH}=n_{NaOH}=0,1(mol)\\ m=0,1.94=9,4(g) \end{array}$

$\begin{array}\\ C_4H_4+H_2\xrightarrow{t^o,Pd/PbCO_3}C_4H_6\\ C_4H_4+3H_2\xrightarrow{t^o,Ni}C_4H_{10} \end{array}$

$\begin{array}{l} a)2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ b)\\ n_{Al}=0,3(mol);n_{H_2SO_4}=0,3(mol)\\ Vì\,\frac{n_{Al}}{2}>\frac{n_{H_2SO_4}}{3}\to Al\,dư\\ n_{Al_2(SO_4)_3}=0,1(mol);n_{H_2}=0,3(mol)\\ m_{Al_2(SO_4)_3}=0,1.342=34,2(g);V_{H_2}=0,3.22,4=6,72(l) \end{array}$

$\begin{array}{l} a)\\ MgO+2HCl\to MgCl_2+H_2O\\ ZnO+2HCl\to ZnCl_2+H_2O\\ b)\\ Hh:\,n_{MgO}=x(mol);n_{ZnO}=y(mol)\\ \to 40x+81y=32,3(1)\\ n_{HCl}=\frac{500}{1000}.2=1(mol)\\ Theo\,PT:\,2x+2y=n_{HCl}=1(2)\\ (1)(2)\to x=0,2;y=0,3\\ \to\begin{cases} m_{MgO}=0,2.40=8(g)\\ m_{ZnO}=0,3.81=24,3(g) \end{cases} \end{array}$

D. Xenlulozơ.

$\begin{array}{l} n_{Mg}=\frac{1,2}{24}=0,05(mol)\\ 200ml=0,2l\\ Mg+2HCl\to MgCl_2+H_2\\ n_{HCl}=2n_{Mg}=0,1(mol);n_{H_2}=n_{Mg}=0,05(mol)\\ \to\begin{cases} V_{H_2}=0,05.22,4=1,12(l)\\ C_{M\,HCl}=\frac{0,1}{0,2}=0,5M \end{cases} \end{array}$

$\begin{array}{l} n_{Zn}=\frac{3,25}{65}=0,05(mol)\\ Zn+H_2SO_4\to ZnSO_4+H_2\\ n_{ZnSO_4}=n_{H_2}=n_{Zn}=0,05(mol)\\ 200ml=0,2l\\ \to\begin{cases} V_{H_2}=0,05.22,4=1,12(l)\\C_{M\,ZnSO_4}=\frac{0,05}{0,2}=0,25M \end{cases} \end{array}$

$\begin{array}{l} Bao\,toan\,KL:\\ m_Y=m_{C_4H_{10}}-5,46=0,15.58-5,46=3,24(g)\\ n_Y=n_{C_4H_{10}}=0,15(mol)\\ Dat\,n_{CO_2}=x(mol);n_{H_2O}=y(mol)\\ \to y-x=0,15(1)\\ n_C=n_{CO_2}=x(mol);n_H=2n_{H_2O}=2y(mol)\\ \to 12x+2y=m_Y=3,24(2)\\ (1)(2)\to x=0,21;y=0,36\\ Bao\,toan\,O:\,n_{O_2}=x+\frac{y}{2}=0,39(mol)\\ \to V=0,39.22,4=8,736(l) \end{array}$

$\begin{array}{l} a)\\ MgO+2HCl\to MgCl_2+H_2O\\ ZnO+2HCl\to ZnCl_2+H_2O\\ b)\\ Hh:\,n_{MgO}=x(mol);n_{ZnO}=y(mol)\\ \to 40x+81y=32,3(1)\\ n_{HCl}=\frac{500}{1000}.2=1(mol)\\ Theo\,PT:\,2x+2y=n_{HCl}=1(2)\\ (1)(2)\to x=0,2;y=0,3\\ \to\begin{cases} m_{MgO}=0,2.40=8(g)\\ m_{ZnO}=0,3.81=24,3(g) \end{cases} \end{array}$

$\begin{array}{l} a)\\ n_{H_2}=\frac{3,36}{22,4}=0,15(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ n_{Fe}=n_{H_2}=0,15(mol)\\ m_{Fe}=0,15.56=8,4(g)\\ b)\\ n_{FeSO_4}=n_{H_2SO_4}=n_{Fe}=0,15(mol)\\ m_{dd\,H_2SO_4}=\frac{0,15.98}{20\%}=73,5(g)\\ \to m_{dd\,spu}=8,4+73,5-0,15.2=81,6(g)\\ \to C\%_{FeSO_4}=\frac{0,15.152}{81,6}.100\%\approx 27,94\% \end{array}$

$\begin{array}{l} CTHH\,B:MgX\\ M_{MgX}=24+M_X=M_{Ca}=40(g/mol)\\ \to M_X=16(g/mol)\\ \to X:O\\ \to B:MgO \end{array}$

$\begin{array}{l} n_{HCl}=\frac{200.3,65\%}{36,5}=0,2(mol)\\ KOH+HCl\to KCl+H_2O\\ n_{KOH}=n_{HCl}=0,2(mol)\\ V_{dd\,KOH}=\frac{0,2}{1}=0,2(l) \end{array}$