The reaction for the burning of one type of match tip is as follows: P4S3(s) + 8 O2(g)---- P4O10(g) + 3 SO2(g) What volume in L of SO2 at STP can be produced by the burning of 1.00 g P4S3 in excess O2?

Question

The reaction for the burning of one type of match tip is as follows: P4S3(s) + 8 O2(g)----> P4O10(g) + 3 SO2(g) What volume in L of SO2 at STP can be produced by the burning of 1.00 g P4S3 in excess O2?

DrBob222 · Answer

P4S3(s) + 8 O2(g)----> P4O10(g) + 3 SO2(g)
mols P4S3 = grams/molar mass = ?
Convert mols P4S3 to mols SO2 using the coefficients in the balanced equation. That's 1 mol P4S3 = 3 mols SO2.
1 mol SO2 at ATP occupies 22.4 L. Convert mols SO2 you have to L.
Post your work if you get stuck.