Ta có: \(n_{MgO}=\dfrac{7}{40}=0,175\left(mol\right)\)

\(n_{H_2SO_4}=0,6.1=0,6\left(mol\right)\)

PT: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

Xét tỉ lệ: \(\dfrac{0,175}{1}< \dfrac{0,6}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{MgSO_4}=n_{H_2SO_4\left(pư\right)}=n_{MgO}=0,175\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,6-0,175=0,425\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{MgSO_4}}=\dfrac{0,175}{0,6}=\dfrac{7}{24}\left(M\right)\\C_{M_{H_2SO_4}}=\dfrac{0,425}{0,6}=\dfrac{17}{24}\left(M\right)\end{matrix}\right.\)

Ta có: \(m_{dd}=300\cdot1,05=315\left(g\right)\) \(\Rightarrow C\%_{Na_2CO_3}=\dfrac{15,9}{315}\cdot100\%\approx5,05\%\)

Mặt khác: \(n_{Na_2CO_3}=\dfrac{15,9}{106}=0,15\left(mol\right)\) \(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)

a) \(n_{NACl}=\frac{5,85}{58,5}=0,1\left(mol\right)\)

    \(n_{AgNO_3}=\frac{34}{170}=0,2\left(mol\right)\)

\(NaCl+AgNO_3\rightarrow AgCl\downarrow+NaNO_3\)

 0,1  \(\rightarrow\)   0,1     \(\rightarrow\)  0,1  \(\rightarrow\)   0,1 (mol)

\(m_{AgCl}=143,5.0,1=14,35g\) 

b) \(V_{dd}=300+200=500\left(ml\right)\)

\(C_M\left(NaNO_3\right)=C_M\left(AgNO_3\right)=\frac{0,1}{0,5}=0,2\left(M\right)\)

a)nNaCl=0,1 mol , nAgNO3=0,2  mol

NaCl+AgNO3---->AgCl+NaNO3

theo pt và theo bài ra: NaCl hết, AgNO3 dư 0,1 mol

=> nAgCl=nNaCl=0,1=>mAgCl=14,35 gam.

b) thể tích sau phản ứng=200+300=500 ml= 0,5 lít 

Nồng độ CMAgNO3=CMNaNO3=0,1/0,5=0,2.

mình làm nhưng nó bị lỗi. Chờ tý để làm lại      

a) PTHH: Na2O + H20 -> 2NaOH 
số mol Na20 = 0,25 (mol) 
=> số mol NaOH = 0,5 mol. 
Nôngd độ mol NaOH = 0,5 / 0,5 = 1 M 
b) PTHH: H2SO4 + 2NaOH -> Na2SO4 + 2H2O 
số mol H2SO4 = 1/2 số mol NaOH = 0,25 mol 
C% H2SO4 = mH2SO4 / m ddH2SO4 . 100% 
=> m ddH2SO4= 122,5 g 
D=m/V => V= 107,5 ml

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có: \(n_{NaOH}=0,3\cdot0,1=0,03\left(mol\right)=n_{HCl\left(p/ứ\right)}\)

+) Trường hợp 1: NaOH còn dư

PTHH: \(Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\uparrow\)

Ta có: \(n_{NaOH\left(dư\right)}=n_{Al}=\dfrac{0,27}{27}=0,01\left(mol\right)\)

\(\Rightarrow n_{HCl}=n_{NaOH\left(p/ứ\right)}=0,02\left(mol\right)\) \(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\)

+) Trường hợp 2: HCl còn dư

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(n_{Al}=0,01\left(mol\right)\) \(\Rightarrow n_{HCl\left(dư\right)}=0,03\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,06\left(mol\right)\) \(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,06}{0,2}=0,3\left(M\right)\) 

\(a.m_{dd}=21,6+400=421,6\left(g\right)\\ b.V_{dd}=400\left(mL\right)\\ c.C_M=\dfrac{\dfrac{21,6}{40}}{0,4}=1,35\left(mol\cdot L^{-1}\right)\)

\(a.m_A=m_{NaOH\left(loãng\right)}=21,6+400=421,6\left(g\right)\)

b,c. thiếu D nhé

Câu 2:

\(m_{H_2SO_4}=39,2.25\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

\(m_{NaOH}=16.30\%=4,8\left(g\right)\Rightarrow n_{NaOH}=\dfrac{4,8}{40}=0,12\left(mol\right)\)

PTHH: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Mol:      0,06         0,12           0,06

Ta có: \(\dfrac{0,1}{1}>\dfrac{0,12}{2}\) ⇒ H₂SO₄ dư, NaOH hết

m_{dd sau pứ} = 39,2+16 = 55,2 (g)

\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,1-0,06\right).98.100\%}{55,2}=7,1\%\)

\(C\%_{ddNa_2SO_4}=\dfrac{0,06.142.100\%}{55,2}=15,43\%\)

Câu 3.

a)\(n_{Ba}=\dfrac{13,7}{137}=0,1\left(mol\right)\)

PTHH: Ba + 2H₂O → Ba(OH)₂ + H₂

Mol:     0,1                    0,1           0,1

m_{dd sau pứ} = 13,7+250.1,008-0,1.2 = 265,5 (g)

\(C\%_{ddBa\left(OH\right)_2}=\dfrac{0,1.171.100\%}{265,5}=6,44\%\)