2a.

601,2=50(ml)nNaOH=20.20100.40=0,1(mol)" role="presentation" style="border:0px; box-sizing:inherit; direction:ltr; display:inline-block; float:none; font-size:18.08px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">601,2=50(ml)nNaOH=20.20100.40=0,1(mol)

0,050,05=1(M)" role="presentation" style="border:0px; box-sizing:inherit; direction:ltr; display:inline-block; float:none; font-size:18.08px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">(M)

\(n_{BaCl_2}=\dfrac{31,2}{208}=0,15mol\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

  0,15        0,15           0,15           0,3

a)\(m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\)

b)\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)

    \(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6}\cdot100=75\left(g\right)\)

c)\(m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\)

  \(m_{ddsau}=31,2+75-34,95=71,25\left(g\right)\)

  \(\Rightarrow C\%_{HCl}=\dfrac{10,95}{71,25}\cdot100\%=15,37\%\)

1.

2K + 2H₂O \(\rightarrow\)2KOH + H₂

n_K=\(\dfrac{19,5}{39}=0,5\left(mol\right)\)

Theo PTHH ta có:

n_K=n_KOH=0,5(mol)

\(\dfrac{1}{2}\)n_K=n_H2=0,25(mol)

m_KOH=56.0,5=28(g)

C% dd KOH=\(\dfrac{28}{19,5+261-0,25.2}.100\%=10\%\)

2.

C% dd muối =\(\dfrac{36}{100+36}.100\%=26,47\%\)

4.

n_Na2CO3.10H2O=\(\dfrac{14,28}{286}=0,05\left(mol\right)\)

Theo CTHH ta có:

n_Na2CO3=n_Na2CO3.10H2O=0,05(mol)

m_Na2CO3=106.0,05=5,3(g)

C% dd=\(\dfrac{5,3}{14,28+200}.100\%=2,47\%\)

Đáp án D

2 mct trong dd ban đầu = 700*12/100 = 84(g)
mct trong dd bão hoà = 84-5 = 79(g)
mdd bão hoà = 700-300-5 = 395 (g)
=> C% = 79*100/395 = 20%

a)

\(C\%_{CuSO_4} = \dfrac{15}{200}.100\% = 7,5\%\\ \)

b)

m HCl = 36,5.2 = 73(gam)

\(C\%_{HCl} = \dfrac{73}{500}.100\% = 14,6\%\)

n_Na = \(\dfrac{6,9}{23}\) =0,3 mol

2Na + 2H₂O ->2 NaOH + H₂

0,3mol ->0,3mol->0,15mol

=>m_NaOH = 0,3 . 40 = 12g

=> m_dd = 6,9 + 50 - 0,15.2 = 56,6 g

=> C% = \(\dfrac{12}{56,6}\).100% = 21,2%

Giúp mk với

1)

\(n_{Na}=\dfrac{m}{M}=\dfrac{6,9}{23}=0,3\left(mol\right)\)

\(n_{H_2O}=\dfrac{m}{M}=\dfrac{50}{18}=\dfrac{25}{9}\left(mol\right)\)

\(\dfrac{n_{Na}}{2}< \dfrac{n_{H_2O}}{2}\)

\(\dfrac{0,3}{2}< \dfrac{\dfrac{25}{9}}{2}\)

=> H₂O dư

\(2Na+2H_2O\xrightarrow[]{}2NaOH+H_2\)

2 : 2 : 2 : 1

0,3 0,3 0,3 0,6 (mol)

\(m_{NaOH}=n.M=0,3.40=1,2\left(g\right)\\ \)

\(m_{dd_{NaOH}}=m_{NaOH}+m_{H_2O}=1,2+50=60,2\left(g\right)\)

\(C\%_{NaOH}=\dfrac{m_{NaOH}}{m_{dd_{NaOH}}}.100\%=\dfrac{1,2}{60,2}.100\%\approx2\%\)