Question

Answer 1

a)

CTHH: S_xO_y

\(\dfrac{m_S}{m_O}=\dfrac{32x}{16y}=\dfrac{40\%}{60\%}\Rightarrow\dfrac{x}{y}=\dfrac{1}{3}\)

=> CTHH: SO₃

b)

CTHH: C_xH_y

\(\dfrac{m_C}{m_H}=\dfrac{12x}{y}=\dfrac{75\%}{25\%}\Rightarrow=\dfrac{x}{y}=\dfrac{1}{4}\)

=> CTHH: CH₄

c)

CTHH: N_xO_y

\(\dfrac{m_N}{m_O}=\dfrac{14x}{16y}=\dfrac{30,435\%}{69,565\%}\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\)

=> CTHH: NO₂

Bài 5: 

a) CTHH: RO

\(NTK_R=80-16=64\left(đvC\right)\)

=> R là Cu

CTHH: CuO

b)

CTHH: T₂CO₃

\(NTK_T=\dfrac{106-60}{2}=23\left(đvC\right)\)

=> T là Na

CTHH: Na₂CO₃

e) 

\(NTK_R=\dfrac{342-96.3}{2}=27\left(đvC\right)\)

=> R là Al

CTHH: Al₂(SO₄)₃

f) \(PTK_T=56x+96y=152\left(đvC\right)\)

Chọn x = 1; y = 1 thỏa mãn

=> CTHH: FeSO₄

Answer 2

Bài 4:

\(a,Đặt.CTTQ:S_xO_y\left(x,y:nguyên,dương\right)\\ Ta.có:\dfrac{\%m_S}{\%m_O}=\dfrac{x.M_S}{y.M_O}\\ \Leftrightarrow\dfrac{40\%}{60\%}=\dfrac{32x}{16y}\\ \Leftrightarrow\dfrac{x}{y}=\dfrac{40\%.16}{60\%.32}=\dfrac{1}{3}\\ \Rightarrow x=1;y=3\\ \Rightarrow CTHH:SO_3\\ b,Đặt.CTTQ:C_aH_b\left(a,b:nguyên,dương\right)\\ Ta.có:\dfrac{\%m_C}{\%m_H}=\dfrac{a.M_C}{b.M_H}\\ \Leftrightarrow\dfrac{75\%}{25\%}=\dfrac{a.12}{b}\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{75\%}{12.25\%}=\dfrac{1}{4}\\ \Rightarrow a=1;b=4\\ \Rightarrow CTHH:CH_4\\ c,Đặt.CTTQ:N_kO_t\left(k,t:nguyên,dương\right)\\ Ta.có:\dfrac{k}{t}=\dfrac{M_O.\%m_N}{M_N.\%m_O}=\dfrac{16.30,435\%}{14.69,565\%}=\dfrac{1}{2}\\ \Rightarrow k=1;t=2\\ \Rightarrow CTHH:NO_2\)