Bạn bổ sung thêm câu hỏi nhé!

\(n_{AgNO3}=0,3\cdot1=0,3\) (mol)

n_HCl =0,5 . 0,5= 0,25 (mol)

_{AgNO3 + HCl -----> \(AgCl\downarrow+HNO3\)}

0,25<--- 0,25--->0,25---->0,25

=>\(n_{AgCl}=0,25\) (mol)

m_AgCl =0,25 . 143,5=35,875(g)

b)\(m_{dd_{ }sau_{ }pứ}\)=300 +500=800(ml)

m_HNO3 =0,25 . 63 =15,75(g)

C_%HNO3= \(\dfrac{15,75}{800}\cdot100\%=1,96\%\)

n_{AgNO3 dư} =0,3 - 0,25 =0,05 (mol)

m_{AgNO3 dư} =0,05 . 143,5=7,175(g)

\(C_{\%AgNO3_{ }dư}\)=\(\dfrac{7,175}{800}\cdot100\%=0,897\%\)

\(n_{NaOH}=0,02.2=0,04\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ a.n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,04}{2}=0,02\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,02}{0,08}=0,25\left(M\right)\\ b.\left[Na^+\right]=\dfrac{0,02.2}{0,02+0,08}=0,4\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,02+0,08}=0,2\left(M\right)\)

Bạn bổ sung thêm số gam dd AgNO₃ nhé.

a)

$K_2SO_4 + BaCl_2 \to BaSO_4 + 2KCl$

b)

$n_{K_2SO_4} = 0,2.2 = 0,4(mol)$
$n_{BaCl_2} = 0,3.1 = 0,3(mol)$

Ta thấy : 

$n_{K_2SO_4} : 1 > n_{BaCl_2} : 1$ nên $K_2SO_4$ dư

$n_{BaSO_4} = n_{BaCl_2} = 0,3(mol)$
$m_{BaSO_4} = 0,3.233 = 69,9(gam)$

c) $n_{K_2SO_4} = 0,4 - 0,3 = 0,1(mol)$

$V_{dd\ sau\ pư} = 0,2 + 0,3 = 0,5(lít)$

$C_{M_{K_2SO_4} } = \dfrac{0,1}{0,5} = 0,2M$
$C_{M_{KCl}} = \dfrac{0,6}{0,5} = 1,2M$

\(200ml=0,2l\\ n_{Na_2CO_3}=0,5.0,2=0,1\left(mol\right)\\ PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ \left(mol\right)........0,1\rightarrow...0,2.......0,2..........0,1.........0,1\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\ b,m_{NaCl}=0,2.58,5=11,7\left(g\right)\\c, V_{ddNaCl}=V_{ddNa_2CO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ C_{M_{NaCl}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

Na₂CO₃  +  2HCl  →  2NaCl  +  H₂O  + CO₂

a) nNa₂CO₃ = 0,2.0,5 = 0,1 mol 

Theo tỉ lệ phản ứng => nHCl = 2nNa₂CO₃ = 0,1.2 = 0,2 mol

=> C_HCl = \(\dfrac{n}{V}\) = \(\dfrac{0,2}{0,2}\)=  1M

b) nNaCl = 2Na₂CO₃ = 0,2 mol

=> mNaCl = 0,2.58,5 = 11,7 gam

c) Thể tích dung dịch sau phản ứng = 0,2 + 0,2 = 0,4 lít

=> C_NaCl = \(\dfrac{0,2}{0,4}\)= 0,5M

\(n_{BaCl_2}=\dfrac{41,6}{208}=0,2\left(mol\right)\)

\(n_{AgNO_3}=\dfrac{17}{170}=0,1\left(mol\right)\)

\(BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+2AgCl\)

\(\dfrac{0,2}{1}>\dfrac{0,1}{2}\) ⇒ BaCl2 dư.

a, \(n_{AgCl}=n_{AgNO_3}=0,1\left(mol\right)\)

\(\Rightarrow m_{AgCl}=0,1.143,5=14,35\left(g\right)\)

b, \(n_{Ba\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgNO_3}=0,05\left(mol\right)\) 

\(C_{M_{Ba\left(NO_3\right)_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)

\(n_{BaCl_2phan/ung}=\dfrac{1}{2}n_{AgNO_3}=0,05\left(mol\right)\)

\(\Rightarrow n_{BaCl_2dư}=0,15\left(mol\right)\rightarrow C_{M\left(BaCl_2\right)}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)