Đáp án + Giải thích các bước giải:

Bài 1:

$n_P=\frac{6,2}{31}=0,2(mol);n_{O_2}=\frac{6,72}{22,4}=0,3(mol)$

$4P+5O_2\xrightarrow{t^o}2P_2O_5$

$\frac{n_P}{4}<\frac{n_{O_2}}{5}\to O_2$ dư

$n_{P_2O_5}=0,5n_P=0,1(mol)$

$m_{P_2O_5}=0,1.142=14,2(g)$

Bài 2:

$a)$

$n_{Fe_2O_3}=\frac{12}{160}=0,075(mol)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$n_{H_2}=3n_{Fe_2O_3}=0,225(mol)$

$V_{H_2}=0,225.22,4=5,04(l)$

$b)$

$n_{Fe}=2n_{Fe_2O_3}=0,15(mol)$

$m_{Fe}=0,15.56=8,4(g)$

Bài 3:

$a)$

$n_{Zn}=\frac{19,5}{65}=0,3(mol)$

$Zn+H_2SO_4\to ZnSO_4+H_2$

$n_{ZnSO_4}=n_{Zn}=0,3(mol)$

$m_{ZnSO_4}=0,3.161=48,3(g)$

$b)$

$n_{H_2}=n_{Zn}=0,3(mol)$

$V_{H_2}=0,3.22,4=6,72(l)$

$c)$

$n_{CuO}=\frac{16}{80}=0,2(mol)$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$n_{CuO}<n_{H_2}\to H_2$ dư

$n_{H_2(dư)}=0,3-0,2=0,1(mol)$

$m_{H_2(dư)}=0,1.2=0,2(g)$

Bài 4:

$a)$

$n_{H_2}=\frac{2,8}{22,4}=0,125(mol)$

$2H_2+O_2\xrightarrow{t^o}2H_2O$

$n_{O_2}=0,0625(mol)$

$V_{kk}=0,0625.5.22,4=7(l)$

$b)$

$n_{H_2O}=n_{H_2}=0,125(mol)$

$m_{H_2O}=0,125.18=2,25(g)$

Đáp án+Giải thích các bước giải:

$1.$

$n_{P}=\dfrac{6,2}{31}=0,2$ $mol$

$n_{O_2}=\dfrac{6,72}{22,4}=0,3$ $mol$

$4P+5O_2\xrightarrow{t^o}2P_2O_5$

Tỉ lệ: $\dfrac{0,2}{4}<\dfrac{0,3}{5}$ $\Rightarrow$ $O_2$ dư

 $\Rightarrow$ $n_{P_2O_5}=\dfrac{1}{2}n_P=\dfrac{1}{2}\times0,2=0,1$ $mol$

 $\Rightarrow$ $m_{P_2O_5}=0,1\times142=14,2g$

$2.$

$a)$ $n_{Fe_2O_3}=\dfrac{12}{160}=0,075$ $mol$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$n_{H_2}=3n_{Fe_2O_3}=3\times0,075=0,225$ $mol$

 $\Rightarrow$ $V_{H_2}=0,225\times22,4=5,04$ $l$

$b)$ $n_{Fe}=2n_{Fe_2O_3}=2\times0,075=0,15$ $mol$

 $\Rightarrow$ $m_{Fe}=0,15\times56=8,4g$

$3.$

$a)$ $n_{Zn}=\dfrac{19,5}{65}=0,3$ $mol$

$Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow$

$n_{ZnSO_4}=n_{Zn}=0,3$ $mol$

 $\Rightarrow$ $m_{ZnSO_4}=0,3\times161=48,3g$

$b)$ $n_{H_2}=n_{Zn}=0,3$ $mol$

 $\Rightarrow$ $V_{H_2}=0,3\times22,4=6,72l$

$c)$ $n_{CuO}=\dfrac{16}{80}=0,2$ $mol$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Tỉ lệ: $\dfrac{0,2}{1}<\dfrac{0,3}{1}$ $\Rightarrow$ $H_2$ dư

 $\Rightarrow$ $n_{H_2(dư)}=0,3-0,2=0,1$ $mol$

 $\Rightarrow$ $m_{H_2(dư)}=0,1\times2=0,2g$

$4.$

$a)$ $n_{H_2}=\dfrac{2,8}{22,4}=0,125$ $mol$

$2H_2+O_2\xrightarrow{t^o}2H_2O$

 $\Rightarrow$ $n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\times0,125=0,0625$ $mol$

 $\Rightarrow$ $V_{O_2}=0,0625\times22,4=1,4l$

 $\Rightarrow$ $V_{kk}=5V_{O_2}=5\times1,4=7l$

$b)$ $n_{H_2O}=n_{H_2}=0,125$ $mol$

 $\Rightarrow$ $m_{H_2O}=0,125\times18=2,25g$

$^{_{_o}}tintinday$