Question

Answer 1

Câu 213 : 

$2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3$
$SO_3 + H_2O \to H_2SO_4$
n H2SO4 = n SO3 = n SO2 = 9,8/22,4 = 7/16 mol

=> C% H2SO4 = (98.7/16) / (100 +80.7/16)    .100% = 31,76%

Answer 2

\(2SO_2+O_2\underrightarrow{t^o}2SO_3\\ SO_3+H_2O\rightarrow H_2SO_4\)

\(n_{H_2SO_4}=n_{SO_3}=n_{SO_2}=\dfrac{9,8}{22,4}=0,4375\left(mol\right)\)

\(m_{H_2SO_4}=0,4375.98=42,875\left(g\right)\)

\(m_{ddH_2SO_4}=0,4375.80+100=135\left(g\right)\)

\(C\%_{ddH_2SO_4}=\dfrac{42,875}{135}.100\%\approx31,76\%\)

Answer 3

Câu 214 : 

$S + O_2 \xrightarrow{t^o} SO_2$
$2SO_2 +O_2 \xrightarrow{t^o,xt} 2SO_3$
$SO_3 + H_2O \to H_2SO_4$