Cho 10.6g Na2CO3 tác dụng hoàn toàn với dung dịch H2SO4 a) Viết phương trình phản ứng sảy ra b) Tính khối lượng muối tạo thành phần sau phản ứng c) Tính thể tích khí CO2 sinh ra ở đktc

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Đáp án:

Ta có: nH2=1,2222,4≈0,05(mol)

a. PTHH: Fe + H2SO4 ---> FeSO4 + H2

b. Theo PT: nFe=nH2=0,05(mol)

=> mFe=0,05.56=2,8(g)

c. Theo PT: nH2SO4=nFe=0,05(mol)

=> 

 

Bạn tham khảo!

$\text{a) Ta có PTHH sau:}$

$\text{Na2CO3+H2SO4$\rightarrow$Na2SO4+H2O+CO2↑}$

$\text{có nNa2CO3=$\frac{m}{M}$=$\frac{10,6}{106}$=0,1 mol}$

$\text{Có n Na2CO3=nNa2SO4=0,1 mol}$

$\rightarrow$ $\text{b) mNa2SO4=n.M=0,1.142=14,2g}$

$\text{Có nCO2=nNa2CO3=0,1 mol}$

$\rightarrow$ $\text{c) V CO2=n.22,4=0,1.22,4=2,24 lít}$

#TRANPHAMGIABAO

 

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