1. C
Ta có:
%$m_{SO3}$ = 71 = $\frac{(32+16.3)n}{98+(32+16.3)n}$ . 100
=> n ≈ 3
2. B
PT: 3H2O + H2SO4.3SO3 ---> 4H2SO4
PƯ: 1 4 (mol)
Ta có:
$m_{dd}$ = $m_{dm}$ + $m_{X ct}$ = 1000 + 1.(98+80.3) = 1338 (g)
C% H2SO4 = $\frac{m_{ct}}{m_{dd}}$ . 100 = $\frac{4.98}{1338}$ . 100
=> C% H2SO4 ≈ 29,3%
Xin câu trả lời hay nhất ạ :3