Question

Lập công thức hợp chất

a,hợp chất A gồm:80%Cu,20%O,mA=80 g

b,hợp chất B gồm 50%S,50%O,mB=64g

Answer 1

\(m_S=\dfrac{50.64}{100}=32\left(g\right)\rightarrow n_S=\dfrac{m}{M}=\dfrac{32}{32}=1\left(mol\right)\)

\(m_O=63-32=32\left(g\right)\rightarrow n_O=\dfrac{32}{16}=2\left(mol\right)\)

\(\rightarrow CTHH:SO_2\)

Answer 2

Lập CTHH :

a) CTTQ : Cu_xO_y

\(\dfrac{64x}{80}\) = \(\dfrac{16y}{20}\) = \(\dfrac{80}{100}\) = \(0,8\)

.\(\dfrac{64x}{80}\) = \(0,8\) ⇒ \(x=\dfrac{0,8.80}{64}\) = \(1\) (\(mol\))

.\(\dfrac{16y}{20}\) = \(0,8\) ⇒ \(y=\dfrac{0,8.20}{16}\)= 1 (\(mol\))

CTHH : CuO

b) CTTQ : S_xO_y

\(\dfrac{32x}{50}\) = \(\dfrac{16y}{50}\) = \(\dfrac{64}{100}\) = \(0,64\)

.\(\dfrac{32x}{50}\) = \(0,64\) ⇒ \(x=\dfrac{0,64.50}{32}\) = \(1\) (\(mol\))

.\(\dfrac{16y}{50}\) = \(0,64\) ⇒ \(y\) = \(\dfrac{0,64.50}{16}\) = \(2\) (\(mol\))

CTHH : SO₂

Chúc bạn học tốt !!!

Answer 3

\(m_{Cu}=\dfrac{80.80}{100}=64\left(g\right)\rightarrow n_{Cu}=\dfrac{m}{M}=\dfrac{64}{64}=1\left(mol\right)\)

\(m_O=80-64=16\left(g\right)\rightarrow n_O=\dfrac{m}{M}=\dfrac{16}{16}=1\left(mol\right)\)

\(\rightarrow CTHH:CuO\)