Đáp án:

Giải thích các bước giải:

$a. Zn + 2HCl$ $\rightarrow$ $ZnCl_2 + H_2(1)$

$n_{Zn}$= $\frac{6,5}{65}$ = $0,1(mol)$

$\text{Ta có: $\dfrac{0,1}{1}$ < $\dfrac{0,25}{1}$}$

$\rightarrow$ $\text{Zn hết, HCl dư.}$

$\text{Từ pt(1):}$ $n_{H_2}$= $0,1(mol)$

$\rightarrow$ $V_{H_2}$= $0,1.22,4= 2,24(l)$

$b.Fe_2O_3 + 3H_2$ $\xrightarrow{t^o}$ $2Fe + 3H_2O(2)$

$n_{Fe_2O_3}$= $\frac{16}{160}$= $0,1(mol)$

$\text{Ta có: $\dfrac{0,1}{1}$ > $\dfrac{0,1}{3}$}$

$\rightarrow$ $\text{$H_2$ hết, $Fe_2O_3$ dư.}$

$n_{Fe_2O_3(dư)}$= $0,1- \frac{1}{30}= \frac{1}{15}(mol)$

$\rightarrow$ $m_{Fe_2O_3(dư)}$= $\frac{1}{15}. 160 \approx 10,67(g)$

Đáp án:

zn+2Hcl→zncl2+h2

a)        nZn=6.5/65=0.1(mol)

         ta có nZn/1=0.1/1<nHcl/2=0.25/2=0.125

           →zn hết ,hcl dư

         theo pt:nH2=nZn=0.1 (mol)

     Vh2=0.1*22.4=2.24

b)    sau pư zn hết ,Hcl dư 

theo pt nHcl=2nZn=0.1*2=0.2( mol )

xin hay nhất

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