How many grams of Fe2O3 are produced from 500.0 g if FeS2 and 400.0 g of oxygen gas?

Question

Consider the following reaction:

4FeS2 + 11O2 -----> 2Fe2O3 + 8SO2

How many grams of Fe2O3 are produced from 500.0 g of FeS2 and 400.0 g of oxygen gas?

a. 35.30
b. 33.17
c. .0302
d. 266.2
e. 1331


Please Show Work!

Answer 1

Molar mass of FeS2 equals 120 g/mol and of O2 is 32 g/mol. First find the moles:

n = m/Mr = 500/120 = 4.17 mol FeS2

n = m/Mr = 400/32 = 12.5 mol O2.

Now, 4 mol FeS2 require 11 mol of O2, so 4.17 mol of FeS2 require:

11*4.17/4 = 11.46 mol of O2. We have more O2, so FeS2 is the limiting reactant. and reacts fully.

4 mol of FeS2 give 2 mol of Fe2O3, so 4.17 mol of FeS2 give:

4.17*2/4 = 2.08 mol Fe2O3

The Fe2O3 has a molar mass of 160, so its mass equals:

m = n*Mr = 2.08*160 = 333.3 g

Answer 2

you may comprehend the proscribing reactant= the reactant that produces smallest volume of product the smallest volume of product is the strategies-blowing product 219 g AlF3/a million * a million mol AlF3/ eighty 3.977 g AlF3 * 6 mol F2/ 4 mol AlF3 * 37.997 g F2/ a million mol F2= 149 g F2 forty 8 g O2/ a million * a million mol O2/ 31.999 g O2 * 6 mol F2/ 3 mol O2* 37.997 g F2 / a million mol F2= 114 g F2 oxygen gasoline is the limitng reactant aluminum flouride is the surplus reactant some is ate up different there to be sure all proscribing reactant is ate up