Question

How many moles of oxygen are needed to completely react with 9.5 grams of sodium

Answer 1

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!

Answer 2

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 $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given volume}}{\text {Molar volume}}$

$\text{Number of moles of sodium}=\frac{9.5g}{23g/mol}=0.413moles$

$4Na+O_2\rightarrow 2Na_2O$

According to stoichiometry:

4 moles of $Na$ combine completely with 1 mole of $O_2$ to give 2 moles of $Na_2O$

Thus 0.413 moles of $Na$ will combine completely with=$\frac{1}{4}\times 0.413=0.103$ moles of $O_2$

Thus 0.103 moles of oxygen are needed to completely react with 9.5 grams of sodium