Step-by-step explanation:
Sodium reacts with oxygen gas in a synthesis reaction to produce sodium oxide. They will react according to the following equation:
4 Na + O₂ ---> 2 Na₂O
9.82 x 10^24 atoms of sodium are reacting 9.5 L of oxygen at STP. First we have to convert both samples intos moles.
According to Avogadro's number there are 6.022 *10^23 atoms of Na in 1 mol of Na. We can use that relationship to find the number of moles of Na.
6.022 * 10^23 atoms of Na = 1 mol of Na
moles of Na = 9.82 x 10^24 atoms * 1 mol/(6.022 * 10^23 atoms)
moles of Na = 16.3 moles
Then we know that 1 mol of any gas at STP occupies 22.4 L. We can convert the 9.5 L of oxygen gas into moles using that conversion factor.
1 mol of O₂ = 22.4 L
moles of O₂ = 9.5 L * 1 mol/(22.4)
moles of O₂ = 0.424 moles
So we found that 0.424 moles of O₂ are reacting with 16.3 moles of Na. One of them is in excess and the other one is the limiting reagent. To find that we will calculate the number of moles of Na₂O that each of them will produce.
4 Na + O₂ ---> 2 Na₂O
If we look at the coefficients of the equation we will see that 4 moles of Na will react with 1 mol of O₂ to give 2 moles of Na₂O. So the molar ratio between Na and Na₂O is 4 to 2. We can use that ratio to find the number of moles of Na₂O that will be produced by 16.3 moles of Na.
4 moles of Na : 2 moles of Na₂O molar ratio
moles of Na₂O = 16.3 moles of Na * 2 moles of Na₂O/(4 moles of Na)
moles of Na₂O = 8.15 moles
We can do something similar with O₂ to find the number of moles of Na₂O that are produced by 0.424 moles of O₂.
1 mol of O₂ : 2 moles of Na₂O molar ratio
moles of Na₂O = 0.424 moles of O₂ * 2 moles of Na₂O/(1 mol of O₂)
moles of Na₂O = 0.848 moles
We found that 0.424 moles of O₂ will produce 0.848 moles of Na₂O (when reacting with excess Na) and 16.3 moles of Na will produce 8.15 moles of Na₂O (when reacting with excess O₂).
So it is clear that O₂ is limiting the reaction and Na is in excess.
O₂ ----> limiting reactant Na ----> excess reactant
So, 0.848 moles of Na₂O are produced, we can convert them into grams using its molar mass
molar mass of Na₂O = 61.98 g/mol
mass of Na₂O = 0.848 moles * 61.98 g/(1 mol)
mass of Na₂O = 52.6 g
Finally we have to find the mass of Na that is in excess.
4 Na + O₂ ---> 2 Na₂O
4 moles of Na will react with 1 mol of O₂. We can find the number of moles of Na that reacted with 0.424 moles of O₂.
4 moles of Na : 1 mol of O₂ molar ratio
moles of Na = 0.424 moles of O₂ * 4 moles of Na/(1 mol of O₂)
moles of Na = 1.696 moles
So, 1.696 moles of Na actually reacted but we originally had 16.3 moles of it. We can find the moles of Na that didn't reacted and finally convert them to grams using its molar mass.
excess Na = 16.3 moles - 1.696 moles
excess Na = 14.604 moles
molar mass of Na = 22.99 g/mol
excess mass of Na = 14.604 moles * 22.99 g/(1 mol)
excess mass of Na = 335.7 g
Answer:
a) 52.6 of Na₂O are produced.
b) 335.7 g of Na will be left over.