C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
What is the total volume of gases remaining when 70 cm3 of ammonia is combusted completely with 50 cm3 of oxygen according to the equation shown?
4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l)
Answer:
Step 1: From the equation deduce the molar ratio of the gases, which is NH3 :O2 :NO or 4:5:4 (water is not included as it is in the liquid state)
Step 2: We can see that oxygen will run out first (the limiting reactant) and so 50 cm3 of O2 requires 4/5 x 50 cm3 of NH3 to react = 40 cm3
Step 3: Using Avogadro's Law, we can say 40 cm3 of NO will be produced
Step 4: There will be of 70-40 = 30 cm3 of NH3 left over
Therefore the total remaining volume will be 40 + 30 = 70 cm3 of gases
Since gas volumes work in the same way as moles, we can use the 'lowest is limiting' technique in limiting reactant problems involving gas volumes. This can be handy if you are unable to spot which gas reactant is going to run out first.Divide the volumes of the gases by the cofficients and whichever gives the lowest number is the limiting reactant
volume of gas (dm3) = amount of gas (mol) x 22.7 dm3 mol-1


To use the gas formula triangle cover the one you want to find out about with your finger and follow the instructions
What is the volume occupied by 3.0 moles of hydrogen at stp ?
Answer:
volume of gas (dm3) = amount of gas (mol) x 22.7 dm3 mol-1
3.0 mol x 22.7 dm3 mol-1= 68 dm3
How many moles are in the following volumes of gases?
Answer 1:
Use the formula:
Answer 2:
Step 1: Convert the volume from cm3 to dm3
960/1000 = 0.960 dm3
Step 2: Use the formula

转载自savemyexams
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