Bài 14:

$a,n_{FeO}=\dfrac{36}{72}=0,2(mol);n_{H_2}=\dfrac{1,5}{2}=0,75(mol)\\ PTHH:FeO+H_2\xrightarrow{t^o}Fe+H_2O\\ \text{Lập tỉ lệ: }\dfrac{0,2}{1}<\dfrac{0,75}{1}\Rightarrow H_2\text{ dư}\\ \Rightarrow V_{H_2(dư)}=n_{H_2(dư)}.24,79=(0,75-0,2).24,79=13,6345(l)\\ b,n_{Fe}=n_{H_2O}=n_{FeO}=0,2(mol)\\ \Rightarrow m_{Fe}=0,2.56=11,2(g);m_{H_2O}=0,2.18=3,6(g)\\ \Rightarrow m_{sp}=m_{Fe}+m_{H_2O}=11,2+3,6=14,8(g)\\ c,PTHH:2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ \Rightarrow n_{Cl_2}=1,5.n_{Fe}=0,3(mol);n_{FeCl_3}=n_{Fe}=0,2(mol)\\ \Rightarrow V_{Cl_2}=0,3.24,79=7,437(l);m_{FeCl_3}=0,2.162,5=32,5(g)$

Bài 15:

$a,n_{C_2H_6O}=\dfrac{9,2}{46}=0,2(mol)\\ PTHH:C_2H_6O+3O_2\to 2CO_2+3H_2O\\ \Rightarrow n_{CO_2}=0,4(mol);n_{H_2O}=0,6(mol)\\ \Rightarrow m_{CO_2}=0,4.44=17,6(g);m_{H_2O}=0,6.18=10,8(g)\\ b,n_{O_2}=n_{H_2O}=0,6(mol)\Rightarrow V_{O_2}=0,6.24,79=14,874(l)\\ \Rightarrow V_{kk}=14,874.5=74,37(l)\\ c,PTHH:2H_2O\xrightarrow{đpdd}2H_2+O_2\\ \Rightarrow n_{H_2}=n_{H_2O}=0,6(mol)\\ \Rightarrow V_{H_2}=0,6.24,79=14,874(l)$

Bài 16:

$a,n_{CO_2}=\dfrac{35,2}{44}=0,8(mol)\\ PTHH:2C_4H_{10}+13O_2\xrightarrow{t^o}8CO_2+10H_2O\\ \Rightarrow n_{C_4H_{10}}=\dfrac{1}{4}n_{CO_2}=0,2(mol)\\ n_{O_2}=\dfrac{13}{8}n_{CO_2}=1,3(mol)\\ \Rightarrow V_{C_4H_{10}}=0,2.24,79=4,958(l)\\ V_{O_2}=1,3.24,79=32,227(l)\\ b,PTHH:2Na+2H_2O\to 2NaOH+H_2\\ n_{H_2O}=\dfrac{10}{8}n_{CO_2}=1(mol) \Rightarrow n_{H_2}=\dfrac{1}{2}n_{H_2O}=0,5(mol)\\ \Rightarrow V_{H_2}=0,5.24,79=12,395(l)$

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