Respuesta :
4 Na + O₂ = 2 Na₂O
4* 23 g Na --------> 16 g O₂
9.5 g Na ------------> ?
Mass of O₂ = 9.5 * 16 / 4 * 23
Mass = 152 / 92
Mass = 1.6521 g of O₂
Molar mass O₂ = 16.0 g/mol
1 mole O₂ ------------ 16.0 g
? mole O₂ ------------ 1.6521 g
mole O₂ = 1.6521 * 1 / 16.0
≈ 0.10325 moles of O₂
hope that helped!
4* 23 g Na --------> 16 g O₂
9.5 g Na ------------> ?
Mass of O₂ = 9.5 * 16 / 4 * 23
Mass = 152 / 92
Mass = 1.6521 g of O₂
Molar mass O₂ = 16.0 g/mol
1 mole O₂ ------------ 16.0 g
? mole O₂ ------------ 1.6521 g
mole O₂ = 1.6521 * 1 / 16.0
≈ 0.10325 moles of O₂
hope that helped!
Answer: 0.103 moles of oxygen
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 Liters 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 volume}}{\text {Molar volume}}[/tex]
[tex]\text{Number of moles of sodium}=\frac{9.5g}{23g/mol}=0.413moles[/tex]
[tex]4Na+O_2\rightarrow 2Na_2O[/tex]
According to stoichiometry:
4 moles of [tex]Na[/tex] combine completely with 1 mole of [tex]O_2[/tex] to give 2 moles of [tex]Na_2O[/tex]
Thus 0.413 moles of [tex]Na[/tex] will combine completely with=[tex]\frac{1}{4}\times 0.413=0.103[/tex] moles of [tex]O_2[/tex]
Thus 0.103 moles of oxygen are needed to completely react with 9.5 grams of sodium
