What volume of steam is produced when 10g of propyne is burnt in excess oxygen at STP
how many moles of propyne in 10g?
The reaction equation will tell you how many moles of H2O you get
each mole occupies 22.4L at STP
The equation is C3H4 + 4O2 ==> 3CO2 + 2H2O
Well, let's see... if we want to know the volume of steam produced when 10g of propyne is burnt, we need to look at the balanced chemical equation for the combustion of propyne.
C3H4 + 5O2 → 3CO2 + 2H2O
According to the equation, for every 1 mole of propyne (C3H4) burnt, we get 2 moles of water (H2O) as a product.
Now, to find the moles of propyne, we need to use its molar mass, which is approximately 40.07 g/mol. So, 10g of propyne is approximately 0.25 moles.
Since the ratio of propyne to water is 1:2, we can multiply the number of moles of propyne by 2 to find the moles of water produced. This gives us 0.5 moles of water.
At STP, 1 mole of gas occupies 22.4 liters of volume. So, 0.5 moles of water would occupy 11.2 liters of volume.
Therefore, when 10g of propyne is burnt, approximately 11.2 liters of steam would be produced.
But hey, keep in mind that this is all theoretical and we're assuming ideal conditions here. In reality, there might be other factors that could affect the volume of steam produced. So, don't go betting your clown shoes on it!
To calculate the volume of steam produced when 10g of propyne is burnt in excess oxygen at STP (Standard Temperature and Pressure), we need to use stoichiometry and the ideal gas law.
1. First, let's write the balanced chemical equation for the combustion of propyne (C₃H₄):
C₃H₄ + 4.5 O₂ → 3 CO₂ + 2 H₂O
2. Determine the molar mass of propyne (C₃H₄):
C: 12.01 g/mol
H: 1.01 g/mol
Molar mass of C₃H₄ = (3 * 12.01 g/mol) + (4 * 1.01 g/mol) = 40.08 g/mol
3. Convert the mass of propyne to moles:
Moles of C₃H₄ = Mass of C₃H₄ / Molar mass of C₃H₄
Moles of C₃H₄ = 10 g / 40.08 g/mol ≈ 0.249 mol
4. Using the stoichiometry of the balanced equation, we can see that for every 0.249 moles of propyne burned, 2 moles of water (H₂O) are produced. Therefore, the moles of water produced is also 0.249 mol.
5. Apply the ideal gas law to calculate the volume of water vapor produced:
PV = nRT
P = Pressure at STP = 1 atm
V = Volume of water vapor (to be determined)
n = Moles of water (H₂O) = 0.249 mol
R = Ideal gas constant = 0.0821 L·atm/(mol·K)
T = Temperature at STP = 273.15 K
Rearranging the equation: V = (nRT) / P
V = (0.249 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
Calculating the volume: V = 5.69 L
Therefore, when 10g of propyne is burned in excess oxygen at STP, it will produce approximately 5.69 liters of steam.