Question

1) How many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7g Fe? Show your work.

Answer 1

1.) is157.7 g
moles Fe = 209.7 g/ 55.847 g/mol=3.75 
the ratio between Fe and CO is 2 : 3 
moles CO = 3.75 x 3 /2 =5.63 
mass CO = 5.63 mol x 28.01 g/mol=157.7 g

2.) is 1.06 moles 
48.7/23 = 2.12 moles sodium 
2.12/2 x 24 = 25.44dm^3 hydrogen = 1.06 moles 
1.06 X 6.02x10^23 = 1.204x10^24 molecules of hydrogen.

3.) is 91.8 
8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced. 8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically. % yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 

Answer 2

Mol Fe = 209.7/56 = 3.744 mol Fe
mol CO = 3.744 * 3/2 (mol ratio) = 5.616 mol CO
molecular weight of CO = 28
5.616 * 28 = 157.28 g CO needed

48.7/23 = 2.11 mol Na
2.11 * 1/2 = 1.055 mol H2
1.055 * 6.023*10^23 = 6.351 x 10^23 molecules of H2

8.3 * 2/2 = 8.3 mol H2O that could be produced
137.1/18 = 7.166 mol actually produced
7.616/8.3 * 100% = 91.75% produced (% yield)