Question

Balance the following reaction and use to answer the questions which follow: S8 + 8O2 -> SO2

A) calculate the # of grams of SO2 are produced by reacting 34.5g S8 with excess O2

B) Calculate the # of grams of O2 that are needed to produce 54.7g of SO2?

Answer 1

S8 + 8O2 ->8 SO2

Answer 2

1 mole of S8 reacts with 8 moles of O2 to form 8 moles of SO2
So the reaction ratio is 1:8:8
S8 (sulfur) = 32,065*8 = 256,520 g/mol
O2 (oxygen) = 15,999*2 = 31,998 g/mol
SO2 (sulfurdioxide) = 32,065 + 31,998 = 64,063 g/mol

Now you have 34,500g S8.
You need to calculate how many moles that is
1 mole S8 = 256,520 gram
So 34,500 gram = 34,500/256,520 = 0,134 moles S8

Now 1 mole S8 reacts with 8 mole O2 to form 8 mole SO2.
So you multiply the weight of 1 mole SO2 by 8
That's 64,063*8 = 512,504 gram SO2

Now you can do the same for part B but reversted
You produce 54,700g of SO2, determine how many moles of SO2 that is. Then use the ratio to calculate how many O2 is used in the reaction.