Question

Answer 1

Câu 5

Gọi khối lượng dd H₂SO₄ 85% là a (g)

Gọi khối lượng dd HNO₃ x% là b (g)

\(m_{H_2SO_4}=\dfrac{85a}{100}=0,85a\left(g\right)\)

\(m_{HNO_3}=\dfrac{bx}{100}=0,01bx\left(g\right)\)

m_{dd sau pư} = a + b (g)

\(C\%_{H_2SO_4}=\dfrac{0,85a}{a+b}.100\%=60\%\)

=> \(\dfrac{a}{b}=\dfrac{12}{5}\)

\(C\%_{HNO_3}=\dfrac{0,01bx}{a+b}.100\%=\dfrac{0,01bx}{\dfrac{12}{5}b+b}.100\%=\dfrac{0,01x}{3,4}.100\%=20\%\)

=> x = 68

Answer 2

Câu 6:

1)

CTHH oxit cao nhất là RO₂

\(\%O=\dfrac{32}{M_R+32}.100\%=53,33\%\)

=> M_R = 28 (g/mol)

=> R là Silic (Si)

2)

\(n_{H_2\left(1\right)}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

PTHH: M_xO_y + yH₂ --t^o--> xM + yH₂O (1)

            \(\dfrac{0,06}{y}\)<--0,06-->\(\dfrac{0,06x}{y}\)

=> \(M_{M_xO_y}=x.M_M+16y=\dfrac{3,48}{\dfrac{0,06}{y}}=56y\left(g/mol\right)\)

=> x.M_M = 42y

=> \(\dfrac{x}{y}=\dfrac{42}{M_M}\) (*)

\(n_{H_2\left(2\right)}=\dfrac{1,008}{22,4}=0,045\left(mol\right)\)

PTHH: 2M + 2nHCl --> 2MCl_n + nH₂

         \(\dfrac{0,09}{n}\)<--------------------0,045

=> \(\dfrac{0,06x}{y}=\dfrac{0,09}{n}\) (**)

(*)(**) => \(\dfrac{2,52}{M_M}=\dfrac{0,09}{n}\)

=> M_M = 28n (g/mol)

- Nếu n = 1 => Loại

- Nếu n = 2 => M_M = 56 (g/mol) --> Fe

- Nếu n = 3 => M_M = 84 (Loại)

Vậy M là Fe

\(\dfrac{x}{y}=\dfrac{42}{56}=\dfrac{3}{4}\) => CTHH: Fe₃O₄