BaCl₂ + H₂SO₄ -> BaSO₄ + 2HCl

n_H2SO4=\(\dfrac{180.15\%}{98}=\dfrac{27}{98}\left(mol\right)\)

n_BaCl2=\(\dfrac{320.10\%}{208}=\dfrac{2}{13}\left(mol\right)\)

VÌ \(\dfrac{27}{98}>\dfrac{2}{13}\) nên H₂SO₄ dư \(\dfrac{155}{1274}\left(mol\right)\)

Theo PTHH ta có:

2n_BaCl2=n_HCl=\(\dfrac{4}{13}\left(mol\right)\)

n_BaSO4=n_BaCl2=\(\dfrac{2}{13}\)

Sau đó tìm mdd rồi tính nhé,lẻ quá ko muốn tính

Câu 1

Ta có \(n_{NaOH}=0,06\left(mol\right)\)

\(n_{H_3PO_4}=0,05\left(mol\right)\)

PT \(NaOH+H_3PO_4\rightarrow NaH_2PO_4+H_2O\) (1)

Ta thấy \(n_{NaOH}>n_{H_3PO_4}\Rightarrow n_{NaOH\left(pu\right)}=n_{H_3PO_4}=0,05\left(mol\right)\)

\(\Rightarrow n_{NaOH\left(du\right)}=0,06-0,05=0,01\left(mol\right)\)

\(NaOH_{\left(du\right)}+NaH_2PO_4\rightarrow Na_2HPO_4+H_2O\) (2)

Ta có chất sau phản ứng gồm \(NaH_2PO_4;Na_2HPO_4\)

Theo (1) \(n_{NaH_2PO_4}=n_{H_3PO_4}=0,05\left(mol\right)\)

Theo (2) \(n_{NaH_2PO_4\left(pu\right)}=n_{NaOH\left(du\right)}=0,01\left(mol\right)\)

\(\Rightarrow n_{NaH_2PO_4\left(du\right)}=0,05-0,01=0,04\left(mol\right)\)

\(n_{Na_2HPO_4}=0,01\left(mol\right)\)

Thể tích dd sau phản ứng là

\(V_{dd}=200+250=450\left(ml\right)=0,45\left(l\right)\)

\(C_M\left(NaH_2PO_4\right)=\dfrac{4}{45}M\)

\(C_M\left(Na_2HPO_4\right)=\dfrac{1}{45}M\)

Câu 2

Ta có \(m_{KOH}=33,6\left(g\right)\Rightarrow n_{KOH}=0,6\left(mol\right)\)

\(m_{H_2SO_4}=49\left(g\right)\Rightarrow n_{H_2SO_4}=0,5\left(mol\right)\)

\(KOH+H_2SO_4\rightarrow KHSO_4+H_2O\) (1)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\) (2)

Ta thấy ở (2) \(n_{KOH}< n_{H_2SO_4}=\dfrac{0,6}{2}< \dfrac{0,5}{1}\)

3) nNaOH = 0,06

2NaOH + H2SO4 —> Na2SO4 + 2H2O

0,06………..0,03

—> mddH2SO4 = 0,03.98/30% = 9,8 gam

—> V = m/D = 8,02 ml

PTHH: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

a) Ta có: \(n_{CuSO_4}=\dfrac{320\cdot20\%}{160}=0,4\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,4\cdot98=39,2\left(g\right)\)

b) Theo PTHH: \(n_{NaOH}=2n_{CuSO_4}=0,8mol\) \(\Rightarrow m_{ddNaOH}=\dfrac{0,8\cdot40}{10\%}=320\left(g\right)\)

c) Theo PTHH: \(n_{Na_2SO_4}=n_{CuSO_4}=0,4mol\) \(\Rightarrow m_{Na_2SO_4}=0,4\cdot142=56,8\left(g\right)\)

Mặt khác: \(m_{dd}=m_{ddNaOH}+m_{ddCuSO_4}-m_{Cu\left(OH\right)_2}=600,8\left(g\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{56,8}{600,8}\cdot100\%\approx9,45\%\) 

Có: \(m_{H_2SO^{ }_4\left(1\right)}=\dfrac{200.10}{100}=20\left(g\right)\)

\(m_{H_2SO_4\left(2\right)}=\dfrac{320.15}{100}=48\left(g\right)\)

\(\Rightarrow\Sigma m_{H_2SO_4}=20+48=68\left(g\right)\)

Có: m_{dd sau}= 200+320=520(g)

\(\Rightarrow C\%=\dfrac{68}{520}.100\%\approx13,077\%\)

cốc 1 chứa 200g dd H2SO4 10%

=> C%= \(\dfrac{m_{ct1}}{m_{dd1}}.100\%=10\%\Rightarrow\dfrac{m_{ct1}}{m_{dd1}}=\dfrac{1}{10}\)

\(\Rightarrow m_{ct1}=20g\)

cốc 2 chứa 320g dd h2so4 15%.

=> C%= \(\dfrac{m_{ct2}}{m_{dd2}}.100\%=15\%\Rightarrow\dfrac{m_{ct2}}{m_{dd2}}=\dfrac{3}{20}\)

\(m_{H_2SO_4.10\%}=200\times10\%=20\left(g\right)\)

\(m_{H_2SO_4.15\%}=320\times15\%=48\left(g\right)\)

\(\Rightarrow\Sigma m_{H_2SO_4}=20+48=68\left(g\right)\)

\(\Sigma m_{ddH_2SO_4}=200+320=520\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{68}{520}\times100\%=13,08\%\)

Vậy khi đổ cốc 1 chứa 200g dd H₂SO₄ 10% vào cốc 2 chứa 320g dd H₂SO₄ 15% ta được dd H₂SO₄ 13,08%

a,\(n_{NaOH}=0,5.1=0,5\left(mol\right);n_{H_2SO_4}=0,5.1=0,5\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:        0,5          0,25             0,25

Ta có: \(\dfrac{0,5}{2}< \dfrac{0,5}{1}\) ⇒ NaOH hết, H₂SO₄ dư

\(m_{Na_2SO_4}=142.0,25=35,5\left(g\right)\)

b,\(C_{M_{Na_2SO_4}}=\dfrac{0,25}{1}=0,25M\)

\(C_{M_{H_2SO_4dư}}=\dfrac{0,5-0,25}{1}=0,25M\)

a/ mdd sau khi hòa tan = 20 + 180 = 200 (g)

C%ddFe(NO3)2 = 20 x 100/200 = 10 (%)

b/ mCuSO4 = n.M = 0.65 x 160 = 104 (g)

C%ddCuSO4 = 104 x 100/320 = 32.5 (%)

a/ Khối lượng dung dịch sau hòa tan: 20+180=200g

C%Fe(NO3)2= 20/200*100%=10%

b/ Khối lượng dd sau hòa tan: 0.65+320=320.65g

C%CuSO4= 0.65/320*100%= 0.203125%