Answer: 313.3 grams
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{585g}{29.88g/mol}=19.58moles[/tex]
1 mole of [tex]Li_2O[/tex] contains = 1 mole of Oxygen atom (O)
Thus 19.58 moles of [tex]Li_2O[/tex] contain =[tex]\frac{1}{1}\times 19.58=19.58moles[/tex] of Oxygen atom (O)
Mass of Oxygen (O)= [tex]moles\times {\text {Atomic mass}}=19.58\times 16=313.3g[/tex]
Thus there are 313.3 grams of O are in 585 g of [tex]Li_2O[/tex].
