Question

Giúp mình với:

Tính thể tích ở đktc của

a)5,6g phân tử C₂H₄

b)0,125 phân tử SO₂

c) 0,75.10²³ phân tử O2

Answer 1

a) \(n_{C_2H_4}=\frac{5,6}{28}=0,2\left(mol\right)\)

\(\Rightarrow V_{C_2H_4}=0,2\times22,4=4,48\left(l\right)\)

b) \(n_{SO_2}=\frac{0,125}{6\times10^{23}}=2,083\times10^{-25}\left(mol\right)\)

\(V_{SO_2}=2,083\times10^{-25}\times22,4=4,66592\times10^{-24}\left(l\right)\)

c) \(n_{O_2}=\frac{0,75\times10^{23}}{6\times10^{23}}=0,125\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,125\times22,4=2,8\left(l\right)\)

Answer 2

\(a.n_{C_2H_4}=\frac{m}{M}=\frac{5,6}{28}=0,2\left(mol\right)\rightarrow V_{C_2H_4}=n.22,4=0,2.22,4=4,48\left(l\right)\)

\(b.n_{SO_2}=\frac{Snt,pt}{N}=\frac{0,125}{6.10^{23}}=2,083333333.10^{-25}\left(mol\right)\rightarrow V_{SO_2}=n.22,4=2,083333333.10^{-25}.22,4\approx4,7.10^{-24}\left(l\right)\)

\(c.n_{O_2}=\frac{Snt,pt}{N}=\frac{0,75.10^{23}}{6.10^{23}}=0,125\left(mol\right)\rightarrow V_{O_2}=n.22,4=0,125.22,4=2,8\left(l\right)\)