What volume is produced when 10g of propyne is burnt in excess oxygen at S. T P
Propyne has the general formula of CnH2n-2 or specifically C3H4.
C3H4 + 4O2 --> 3CO2 + 2H2O
mols C3H4 = grams/molar mass = 10/40 = 0.25
Therefore, 0.25 mols C3H4 will produce 3*0.25 mols CO2 plus 2*0.25 mols H2O for a total of 0.75 + 0.50 = 1.25 mols of gas. Then each mol gas @ STP occupies 22.4 L.
Well, if propyne is being burnt, I bet it's trying to make a big entrance like a fiery clown at a circus! At STP (Standard Temperature and Pressure), we can assume that one mole of any gas occupies 22.4 liters.
Now, to find the volume produced, we need to calculate the number of moles of propyne.
The molar mass of propyne (C3H4) is around 40 g/mol. So, 10 g of propyne is roughly 0.25 moles (10 g / 40 g/mol).
Since propyne is being burnt in excess oxygen, it will produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation is:
2 C3H4 + 9 O2 -> 6 CO2 + 4 H2O
Based on the equation, for every 2 moles of propyne burned, you get 6 moles of carbon dioxide. So, 0.25 moles of propyne would produce (6/2) x 0.25 = 0.75 moles of carbon dioxide.
Since 1 mole of any gas is 22.4 liters at STP, 0.75 moles of carbon dioxide would be approximately 0.75 x 22.4 = 16.8 liters.
So, to answer your question, when 10g of propyne is burnt in excess oxygen at STP, approximately 16.8 liters of carbon dioxide are produced. Just make sure to keep your distance from the fire-breathing clown!
To determine the volume of gas produced when 10g of propyne (C₃H₄) is burnt in excess oxygen at standard temperature and pressure (STP), we need to use the concept of stoichiometry and the ideal gas law.
1. Determine the balanced chemical equation for the combustion of propyne:
C₃H₄ + O₂ → CO₂ + H₂O
2. Calculate the number of moles of propyne:
Using the molar mass of propyne (C₃H₄ = 40.06 g/mol), we can find the number of moles:
10 g / 40.06 g/mol = 0.250 mol
3. Determine the stoichiometric ratio between propyne and its gaseous products:
From the balanced equation, we can see that for every mol of propyne combusted, we obtain one mol of carbon dioxide (CO₂) and one mol of water (H₂O).
4. Calculate the moles of carbon dioxide produced:
Since the stoichiometric ratio is 1:1, we have 0.250 mol of carbon dioxide.
5. Convert moles of carbon dioxide to volume at STP:
One mole of any gas occupies 22.4 liters at STP. Therefore, 0.250 mol of carbon dioxide will occupy:
0.250 mol × 22.4 L/mol = 5.6 L
Therefore, when 10g of propyne is burnt in excess oxygen at STP, it will produce 5.6 liters of carbon dioxide.
Where is the answer
how many moles involved?
1 mole occupies 22.4L at STP