mik gửi nha:

chúc bn hok tốt.

@nguyensyannamDH7qc

Bài 1:

$n_{H_2SO_4}=\dfrac{180.15\%}{100\%.98}\approx 0,275(mol)\\ n_{BaCl_2}=\dfrac{320.10\%}{100\%.208}\approx 0,15(mol)\\ PTHH:BaCl_2+H_2SO_4\to BaSO_4\downarrow+2HCl\\ \text{Lập tỉ lệ: }\dfrac{0,275}{1}>\dfrac{0,15}{2}\Rightarrow H_2SO_4\text{ dư}\\ \Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,15(mol);n_{HCl}=2n_{BaCl_2}=0,3(mol)\\ n_{H_2SO_4(dư)}=0,275-0,15=0,125(mol)\\ \Rightarrow m_{BaSO_4}=0,15.233=34,95(g);m_{CT_{HCl}}=0,3.36,5=10,95\\ m_{CT_{H_2SO_4(dư)}}=0,125.98=12,25(g)\\ \Rightarrow m_{\text{dd sau p/ứ}}=180+320-34,95=465,05(g)\\ \Rightarrow \begin{cases} C\%_{H_2SO_4(dư)}=\dfrac{12,25}{465,05}.100\%=2,63\%\\ C\%_{HCl}=\dfrac{10,95}{465,05}.100\%=2,35\% \end{cases}$

Bài 2:

$n_{Ba(OH)_2}=0,4.0,2=0,08(mol);n_{H_2SO_4}=0,3.0,25=0,075(mol)\\ PTHH:Ba(OH)_2+H_2SO_4\to BaSO_4\downarrow+2H_2O\\ \text{Lập tỉ lệ: }\dfrac{0,08}{1}>\dfrac{0,075}{1}\Rightarrow Ba(OH)_2\text{ dư}\\ \Rightarrow n_{BaSO_4}=n_{H_2SO_4}=0,075(mol)\\ \Rightarrow m_{\downarrow}=m_{BaSO_4}=0,075.233=17,475(g)$