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Cho 4g CuO vào dung dịch axit clohidric 10% thì phản ứng vừa đủ.
a/ Viết PThh
b/ tính khối lượng muối sinh ra
c/ tính nồng độ % dung dịch sau phản ứng

Sagot :

Answer:

Explanation:

a. CuO+ 2HCl⇒CuCl2+ H2O

b. [tex]n_{CuO}[/tex]= [tex]\frac{4}{80}[/tex]= 0,05 (mol)

⇒[tex]n_{CuCl2}[/tex]= [tex]n_{CuO}[/tex]=0,05 mol

⇒[tex]m_{CuCl2}[/tex]= 0,05×135=6,75 (g)

c. [tex]n_{HCl}[/tex]=2× [tex]n_{CuO}[/tex]=0,1 (mol)

⇒[tex]m_{HCl}[/tex]= 0,1×36,5= 3,65 (g)

⇒[tex]m_{dd HCl}[/tex]= [tex]\frac{m_{HCl}}{10}[/tex]×100=36,5 (g)

⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=[tex]\frac{m_{CuCl2} }{m_{dd HCl+ m_{CuO} } }[/tex]×100=[tex]\frac{6,75}{36,5+4}[/tex] ×100≈ 16,67%

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