bai1 cho m gam bot sat vao dd axit sunfuric loang du phan ung hoan toan tao ra 6,72lit khi hidro gia tri cua m la/bai2 suc 1,6 gam khi so3 vao nuoc du thu duoc dd chua m gam h2so4 gia tri cua m la
bai1 cho m gam bot sat vao dd axit sunfuric loang du phan ung hoan toan tao ra 6,72lit khi hidro gia tri cua m la/bai2 suc 1,6 gam khi so3 vao nuoc du thu duoc dd chua m gam h2so4 gia tri cua m la
`1.`
` nH2 = 0,3 (mol)`
` Fe + H_2SO_4 -> FeSO_4 + H_2`
` => nFe = nH_2 = 0,3 (mol)`
` => mFe = 16,8(g)`
`2.`
` nSO_3 = 0,02 (mol)`
` SO_3 + H_2O -> H_2SO_4`
` => nH_2SO_4 = nSO_3 = 0,02 (mol)`
`=> mH_2SO_4 = 0,02.98=1,96(g)`
1)
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3\,{\text{mol = }}{{\text{n}}_{Fe}}\)
\( \to m = {m_{Fe}} = 0,3.56 = 16,8{\text{ gam}}\)
2)
Phản ứng xảy ra:
\(S{O_3} + {H_2}O\xrightarrow{{}}{H_2}S{O_4}\)
Ta có:
\({n_{S{O_3}}} = \frac{{1,6}}{{32 + 16.3}} = 0,02{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,02.98 = 1,96{\text{ gam}}\)