Cho 10,05 gam hỗn hợp X chứa MgO và ZnO tác dụng vừa đủ với dung dịch H2SO4 loãng, thu được một dung dịch chứa 26,05 gam hỗn hợp muối. Tính thành phần % theo khối lượng mỗi oxit trong hỗn hợp X. Cho Mg=24, Zn=654

2 câu trả lời

Giải thích các bước giải:

MgO+H2SO4→MgSO4+H2O

ZnO+H2SO4→ZnSO4+H2O

mhh=40a+81b=10,05(1)

mMuối=120a+161b=26,05(2)

(1)(2)

a=0,15

b=0,05

%mMgO=0,15.4010,05.100%=59,7%

%mZnO=40,3%

Bạn tham khảo:

$MgO+H_2SO_4 \to MgSO_4+H_2O$

$ZnO+H_2SO_4 \to ZnSO_4+H_2O$

$m_{hh}=40a+81b=10,05(1)$

$m_{Muối}=120a+161b=26,05(2)$

$(1)(2)$

$a=0,15$

$b=0,05$

$\%m_{MgO}=\dfrac{0,15.40}{10,05}.100\%=59,7\%$

$\%m_{ZnO}=40,3\%$

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