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.
attemtp at solution
volume of oxygen= .209476*50L=10.5L
How do i go from volume of oxygen to moles of oxygen the density of air is given not the density of oxygen.
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To calculate the number of moles of oxygen, we need to convert the volume of oxygen to moles by using the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume of the gas (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L atm/mol K)
T = temperature (in Kelvin)
Since we are given the volume of oxygen in liters, we can directly use it in the equation. At STP, the pressure is 1 atm and the temperature is 273.15 K (0 degrees Celsius).
So, we have:
1 atm * 10.5 L = n * 0.0821 L atm/mol K * 273.15 K
Solving for n, we get:
n = (1 atm * 10.5 L) / (0.0821 L atm/mol K * 273.15 K)
n ≈ 0.405 moles
Therefore, the number of moles of oxygen present in 10.5 liters of air at STP is approximately 0.405 moles.
Now, to calculate the number of grams of oxygen, we can use the molar mass of oxygen, which is approximately 32 g/mol.
So, the number of grams of oxygen is:
Grams of oxygen = moles of oxygen * molar mass of oxygen
= 0.405 moles * 32 g/mol
≈ 12.96 grams
Therefore, there are approximately 12.96 grams of oxygen in 10.5 liters of air at STP.