Question

2 C2H2(g) + 5 02 (9)(g) – 4 CO2(g) + 2 H2O(g)2What volume of carbon dioxide gas (in mL) is produced when 275 mL of acetylene gas undergoescomplete combustion? Assume all reactant and product volumes are measured at SATP?

Answer 1

In order to answer this question we need to know that in SATP the molar volume is 24.5 L/mol, so this means that for every mole that in have in Standard Ambient Temperature and Pressure, we will have 24.5 Liters of it.

In our question we have this reaction:

2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O

We already have the volume of C2H2, which is 275 mL or 0.275 L, now we need to find the number of moles

24.8 L = 1 mol

0.275 L = x moles

x = 0.011 moles of C2H2

Now if you check the molar ratio for C2H2 and CO2, you will notice that we have 2x more moles of CO2 than C2H2, this means that the molar ratio is 2:4, for every 2 moles of C2H2 in a reaction we will end up with 4 moles of CO2 as product.

2 C2H2 = 4 CO2

0.011 C2H2 = x CO2

x = 0.022 moles of CO2

Now we will use again that molar volume information to find out what is the volume of carbon dioxide:

24.8 L = 1 mol

x Liters = 0.022 moles

x = 0.546 Liters or 546 mL