Anh làm rồi em nhé

\(m_{H_2SO_425\%}=\dfrac{150.25}{100}=37,5\left(g\right)\)

Gọi m dd H2SO4 10% là x (g) 

Ta có : m dd H2SO4 15% = 150+x (g) 

\(m_{H_2SO_415\%}=\dfrac{10x}{100}+37,5=0,1x+37,5\left(g\right)\)

\(15=\dfrac{0,1x+37,5}{x+150}.100\)

\(\rightarrow0,15=\dfrac{0,1x+37,5}{x+150}\)

\(\rightarrow0,05x=15\)

\(\rightarrow x=300\)

Vậy m dd H2SO4 10% là : 300g

Gọi mdd H2SO4 10% = x (g)

Ta có: mH2SO4 25% = \(\frac{150.25}{100}\) = 37,5 g

mdd H2SO4 15% = x + 150

mH2SO4 15% = \(\frac{10x}{100}\) +37,5= 0,1x + 37,5

Ta có: C% dd sau p.ứng =\(\frac{m_{ct}}{m_{dd}}.100\)

⇔⇔ 15 =\(\frac{0,1x+37,5}{x+150}.100\)

⇔⇔ 0,15 = \(\frac{0,1x+37,5}{x+150}\)

⇔⇔ 0,15x + 22,5 = 0,1x + 37,5

⇔⇔ 0,05x = 15

⇔⇔ x = 300g = mdd H2SO4 10%

Gọi m_{dd H2SO4 10%} = x (g)

Ta có: m_{H2SO4 25%} = \(^{\frac{150}{25}.100}\) = 37,5 g

m_{dd H2SO4 15%} = x + 150

m_{H2SO4 15%} = \(\frac{10.x}{100}\) + 37,5 = 0,1x + 37,5

Ta có: C% _{dd sau p.ứng} = \(\frac{m_{ct}}{m_{dd}}.100\) 

\(\Leftrightarrow\) 15 = \(\frac{0,1x+37,5}{x+150}.100\)

\(\Leftrightarrow\) 0,15 = \(\frac{0,1x+37,5}{x+150}\)

\(\Leftrightarrow\) 0,15x + 22,5 = 0,1x + 37,5

\(\Leftrightarrow\) 0,05x = 15

\(\Leftrightarrow\) x = 300g = m_{dd H2SO4 10%}

Câu 1 : 

Gọi $n_{SO_3} = a(mol)$

$SO_3 + H_2O \to H_2SO_4$

Sau khi pha : 

$m_{dd} = 80a + 200(gam)$
$m_{H_2SO_4} = 98a + 200.9,8\% = 98a + 19,6(gam)$

Suy ra : $C\%_{H_2SO_4} = \dfrac{98a + 19,6}{200 + 80a}.100\% = 49\%$
$\Rightarrow a = 1,333(mol)$
$m_{SO_3} = 1,333.80 = 106,64(gam)$

Gọi $m_{oleum} = a(gam) ;  m_{H_2SO_4} = b(gam)$
Ta có : 

Sau khi trộn  :

$m_{oleum} = a + b(gam)$
$m_{SO_3} = a.71\% = 0,71a(gam)$

$\Rightarrow \%SO_3 = \dfrac{0,71a}{a + b}.100\% = 62\%$

$\Rightarrow a + b = 0,4402a$

$\Rightarrow \dfrac{a}{b} = \dfrac{1}{1 - 0,4402} = 1,78$

1.

\(m_{H_2SO_4\left(9.8\%\right)}=200\cdot9.8\%=19.6\left(g\right)\)

\(n_{H_2SO_4}=\dfrac{19.6}{98}=0.2\left(mol\right)\)

\(TC:n_{SO_3}=a\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(a................a\)

\(m_{H_2SO_4\left(tt\right)}=98a\left(g\right)\)

\(m_{H_2SO_4\left(tổng\right)}=19.6+98a\left(g\right)\)

\(m_{dd_{H_2SO_4}}=200+80a\left(g\right)\)

\(C\%_{H_2SO_4}=\dfrac{19.6+98a}{200+80a}\cdot100\%=49\%\)

\(\Rightarrow a=\dfrac{4}{3}\)

\(m_{SO_3}=\dfrac{4}{3}\cdot80=106.67\left(g\right)\)

1. \(n_{H_2SO_4\left(98\%\right)}=\dfrac{30.1,84.98\%}{98}=0,552\left(mol\right)\)

=>\(V_{H_2SO_4\left(1M\right)}=\dfrac{0,552}{1}=0,552\left(l\right)\)

\(a) V_A = 2(lít) \to V_B = 3(lít)\\ \Rightarrow V_{dd} = 2 + 3 = 5(lít)\\ n_{H_2SO_4\ trong\ C} = 0,2.2 + 0,5.3 = 1,9(mol)\\ C_{M_{H_2SO_4}} = \dfrac{1,9}{5} = 0,38M\)

\(m_{HCl\left(1\right)}=600.2,5\%=15g\\ m_{HCl\left(2\right)}=400.15\%=60g\\ C\%_{HCl.sau}=\dfrac{75}{1000}\cdot100\%=7,5\%\)