Đáp án:
1/
$V_{H_2}=4,48l.$
$m_{NaOH}=16g.$
2/
$b,V_{H_2}=4,48l.$
$c,m_{H_2}(dư)=0,1g.$
3/
$a,m_{HCl}(dư)=3,65g.$
$b,V_{H_2}=4,48l.$
4/
$a,V_{H_2}=6,72l.$
$b,m_{Fe}=11,2g.$
Giải thích các bước giải:
1/
$PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$
$n_{Na}=\dfrac{9,2}{23}=0,4mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{Na}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$Theo$ $pt:$ $n_{NaOH}=n_{Na}=0,4mol.$
$⇒m_{NaOH}=0,4.40=16g.$
2/
$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$
$b,n_{Zn}=\dfrac{13}{65}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$c,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$n_{CuO}=\dfrac{12}{80}=0,15mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,15}{1}<\dfrac{0,2}{1}$
$⇒n_{H_2}$ $dư.$
$⇒n_{H_2}(dư)=0,2-\dfrac{0,15.1}{1}=0,05mol.$
$⇒m_{H_2}(dư)=0,05.2=0,1g.$
3/
$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$
$n_{Zn}=\dfrac{13}{65}=0,2mol.$
$n_{HCl}=\dfrac{18,25}{36,5}=0,5mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,2}{1}<\dfrac{0,5}{2}$
$⇒n_{HCl}$ $dư.$
$⇒n_{HCl}(dư)=0,5-\dfrac{0,2.2}{1}=0,1mol.$
$⇒m_{HCl}(dư)=0,1.36,5=3,65g.$
$b,Theo$ $pt:$ $n_{ZnCl_2}=n_{Zn}=0,2mol.$
$⇒m_{ZnCl_2}=0,2.136=27,2g.$
$c,Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
4/
$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$
$n_{Zn}=\dfrac{19,5}{65}=0,3mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,3mol.$
$⇒V_{H_2}=0,3.22,4=6,72l.$
$b,PTPƯ:Fe_2O_3+3H_2\xrightarrow{t^o} 2Fe+3H_2O$
$n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,12}{1}>\dfrac{0,3}{3}$
$⇒n_{Fe_2O_3}$ $dư.$
$\text{⇒Tính theo}$ $n_{H_2}$
$Theo$ $pt:$ $n_{Fe}=\dfrac{2}{3}n_{H_2}=0,2mol.$
$⇒m_{Fe}=0,2.56=11,2g.$
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