Calculate the number of grams of reactant left over when 28.35 grams of silver reacts with 50.0 liters of air at STP which is 20.9476 % oxygen by volume. The density of air at STP is 1.292 g/L.
attemt at a solution:
m = dv = 64.6 g of air
moles silver =28.35/108=0.26
moles of air : n= PV/RT = (1atm)(50L)/(.0821)(298K)=2mol
moles of air left= 2 - 0.26 =1.74 mol
iam stuck on this need some help please
Its the oxygen reacting with silver, not air; therefore, I think I would get the oxygen out of the air first. If the air is that percent oxygen, I think I would multiply 50.0 L air x 0.209476 = ?? L oxygen. Convert L oxygen to mols oxygen and go from there.
Ag + O2 ==> Ag2O
You have mols Ag. You have mols oxygen. Determine the limiting reagent. That reagent will be used completely. The other will have some remaining. You can determine the amount used by the amount of Ag2O formed. Then subtract to see how much remains.
To determine the number of grams of reactant left over, you need to compare the moles of the reactant (silver) with the moles of the limiting reactant (air).
First, you correctly calculated the moles of silver:
moles of silver = mass / molar mass = 28.35 g / 108 g/mol = 0.26 mol
Next, let's calculate the moles of air. You used the ideal gas law equation, which is correct:
n = PV / RT
Given:
P = 1 atm (since it is STP)
V = 50.0 L
R = 0.0821 L·atm/(mol·K) (gas constant)
T = 298 K
n = (1 atm) * (50.0 L) / (0.0821 L·atm/(mol·K) * 298 K)
n = 1.9967 mol ≈ 2 mol
Now, to find the moles of air molecules remaining after the reaction with silver:
moles of air left = moles of air - moles of silver
= 2 mol - 0.26 mol
= 1.74 mol
Finally, we can calculate the mass of air left in grams.
mass of air left = moles of air left * molar mass of air
= 1.74 mol * 64.6 g/mol
= 112.404 g
Therefore, there are approximately 112.4 grams of air left over when 28.35 grams of silver reacts with 50.0 liters of air at STP.