Cho 10,2 gam Al2O3 tác dụng vừa đủ với 200ml dung dịch axit sunfuric a. Viết phương trình hóa học b. Tính khối lượng muối tạo thành c. Tính nồng độ mol của dung dịch axit sunfuric đã dùng.

2 câu trả lời

a) PTHH là:

        Al2O3+3H2SO4→Al2(SO4)3+3H2O

b)nAl2O3=m/M=10,2/102=0,1 (mol)

PTHH          Al2O3+3H2SO4→Al2(SO4)3+3H2O

                        0,1        0,3              0,1                        mol

mAl2(SO4)3=n.M=0,1.342=34,2 (g)

c)Đổi 200ml=0,2 l

CMH2SO4=n/V=0,3/0,2=1,5 (M)

`a)`

`Al_2O_3+ 3H_2SO_4→ Al_2(SO_4)_3 + 3H_2O` 

`0,1→ 0,3 → 0,1(mol)`

`b)` 

`n_{Al_2O_3}= {10,2}/{102}= 0,1(mol)` 

`m_{Al_2(SO_4)_3}= 0,1 .342= 34,2(g)` 

`c)`

`200ml= 0,2l` 

`C_{M_{H_2SO_4}}= {0,3}/{0,2}= 1,5(M)`

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