Cho Zn tác dụng 500 ml dd axit Clohiđric thu đc 4,48 lít khí hiđro ở đktc. Tính: a) Khối lượng Zn đã phản ứng b) Nồng độ mol của dd axit c) Tính khối lượng muối có trong dd thu được sau phản ứng

2 câu trả lời

Đáp án:

a)zn+2hcl--->zncl2+h2

n zn= n h2=4,48/22,4=0,2(mol)

=>m zn=0,2.65=13(g)

b)n hcl= 2.n h2=2.0,2=0,4(mol)

CM hcl=0,4/0,5=0,8M

c)n zncl2=n h2=0,2 (mol)

=>m zncl2=0,2.136=27,2(g)

 

Bạn tham khảo!

$\text{Ta có PTHH sau:}$

$\text{Zn+2HCl$\rightarrow$ZnCl2+H2}$

$\text{Số mol khí H2 là:}$

$\text{n=$\frac{V}{22,4}$=$\frac{4,48}{22,4}$=0,2 mol}$

$\rightarrow$ $\text{nZn=0,2 mol}$

$\rightarrow$ $\text{a) mZn=n.M=0,2.65=13g}$

$\text{có nHCl=2nH2=0,2.2=0,4 mol}$

$\rightarrow$ $\text{b) CM HCl=$\frac{n}{V}$=$\frac{0,4}{0,5}$=0,8M}$

$\text{Có nZnCl2=0,2 mol}$

$\rightarrow$ $\text{c) mZnCl2=n.M=0,2.136=27,2g}$

#TRANPHAMGIABAO

 

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