Question

 

Co 10,08 l khi CO2. Tinh so mol,khoi luong

Answer 1

	\(V_{CO_2}=10,08l\)
	\(n_{CO_2}=?\left(mol\right)\) \(m_{CO_2}=?\left(g\right)\)

\(n_{CO_2}=\frac{V_{CO_2}}{22,4}=\frac{10,08}{22,4}=0,45\left(mol\right)\)

\(M_{CO_2}=1\times12+2\times32=44\) (g/mol)

\(m_{CO_2}=n_{CO_2}\times M_{CO_2}=0,45\times44=19,8\left(g\right)\)

Answer 2

n_CO² = V/22,4=10,08/22,4=0,45(mol)
M_CO² = 44g/mol
m_CO² = n x M = 0,45 x 44 =19,8

Answer 3

nCO2 = V/22,4=10,08/22,4=0,45(mol)
MCO2 = 44g/mol
mCO2 = n x M = 0,45 x 44 =19,8 (g)