Đáp án:

Bài 1: b/ $⇒C_{M(H_2SO_4)}=0,625M$

 Bài 2:

a/ $C_{M(H_2SO_4)}=\dfrac{122}{245}\ mol$

b/ $⇒V_{dd\ KOH}≈0,996\ lít$

c/ $⇒m_{dd\ BaCl_2}≈99,6g$

Giải thích các bước giải:

 Bài 1: 

a/ $S+O_2\xrightarrow{t^o}SO_2\\2SO_2+O_2\xrightarrow{t^o,V_2O_5}2SO_3\\SO_3+H_2O\to H_2SO_4$

b/ $n_S=\dfrac{16}{32}=0,5\ mol$

Bảo toàn nguyên tố S, ta có:

$n_S=n_{S(trong\ H_2SO_4)}⇒n_{H_2SO_4}=0,5 mol$

$⇒C_{M(H_2SO_4)}=\dfrac{0,5}{0,8}=0,625M$

Bài 2: 

$n_{SO_3}=\dfrac{11,2}{22,4}=0,5\ mol$

$m_{dd\ H_2SO_4}=200.1,5=300g⇒m_{H_2SO_4}=\dfrac{300.65}{100}=195g$

$⇒n_{H_2SO_4\ ban\ đầu}=\dfrac{195}{98}\ mol$

PTHH

$SO_3+H_2O\to H_2SO_4\\0,5\hspace{3cm}0,5$

$⇒n_{H_2SO_4}=n_{H_2SO_4\ ban\ đầu}+n_{H_2SO_4(1)}=\dfrac{195}{98}+0,5=\dfrac{122}{49}\ mol$

a/ $C_{M(H_2SO_4)}=\dfrac{122}{49.5}=\dfrac{122}{245}\ mol$

b/ Trong 100 ml dung dịch A

$n_{H_2SO_4}=\dfrac{122}{245}.0,1=\dfrac{122}{2450}\ mol$

$2KOH+H_2SO_4\to K_2SO_4+2H_2O$

$⇒n_{KOH}=2.n_{H_2SO_4}=2.\dfrac{122}{2450}=\dfrac{244}{2450}$

$⇒V_{dd\ KOH}=\dfrac{244}{2450}:0,1≈0,996\ lít$

c/

Trong 50 ml dung dịch A

$n_{H_2SO_4}=\dfrac{122}{245}.0,05=\dfrac{61}{2450}\ mol$

$BaCl_2+H_2SO_4\to BaSO_4↓+2HCl$

$n_{BaCl_2}=n_{H_2SO_4}=\dfrac{61}{2450}\\⇒m_{BaCl_2}=\dfrac{61}{2450}.208=\dfrac{6344}{1225}g$

$⇒m_{dd\ BaCl_2}=\dfrac{6344}{1225}.\dfrac{100}{5,2}≈99,6g$