Question

1) How many grams of CO are needed with an excess of Fe2O3 to produce 35.0 g Fe?

Answer 1

1. 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. 2Na + 2H2O --> 2NaOH + H2atomic weight of sodium is 23             48.7/23 = 2.12 moles  of sodium 
2.12/2 x 24 = 25.44dm^3 hydrogen = 1.06 moles

so multiply it by Avogadro's number1.06 X 6.02x10^23 = 1.204x10^24 molecules of hydrogen