Đáp án + Giải thích các bước giải:

 `SO_3 + H_2O->H_2SO_4`

a)

`100ml = 0,1l`

`n_{SO_3}=m/M=2/(80)= 0,025 (mol)`

Theo PT: `n_{H_2SO_4}=n_{SO_3}= 0,025 (mol)`

`=> C_{M_{H_2SO_4}}=n/V=(0,025)/(0,1)= 0,25 (M)`

b)

`m_{H_2O}=D xx V=1 xx 100= 100 (g)`

`=> m_{\text{dd sau pứ}}=2 + 100= 102 (g)`

`=> \text{C%}_{H_2SO_4}=(m_{ct})/(m_{dd}) xx 100\text{%}=(0,025 xx 98)/(102) xx 100\text{%}= 2,4\text{%}`

\(\begin{array}{l}
a)\\
n_{SO_3}=\frac{m}{M}=\frac{2}{80}=0,025(mol)\\
100ml=0,1l\\
SO_3+H_2O\to H_2SO_4\\
Theo\,PT:\,n_{H_2SO_4}=n_{SO_3}=0,025(mol)\\
\to C_{M\,H_2SO_4}=\frac{n}{V_{dd}}=\frac{0,025}{0,1}=0,25M\\
b)\\
m_{H_2O}=V.D=100.1=100(g)\\
\to m_{dd\,H_2SO_4}=2+100=102(g)\\
\to C\%_{H_2SO_4}=\frac{m_{ct}}{m_{dd}}.100\%=\frac{0,025.98}{102}.100\%\approx 2,4\%
\end{array}\)