Question

1. ^{Thefollowingequationdescribestherustingofiron.Note:rustisFe}2^O3(s)

^Fe(s) ^{+ O}2(g) ^Fe2^O3(s)

a. Balance the above equation.

b. If you start with 100.0 grams of iron, how many moles of iron do you have?

c. If you start with 100.0 grams of rust, how many moles of rust do you have?

d How many moles of rust could be formed from 5.00 moles of iron?

e. How many moles of rust could be formed from 5.00 moles of oxygen?

f. If 50. moles of rust formed, how many grams of rust does this represent?

g. If 10.0 moles of oxygen is consumed in the reaction, how many grams of oxygen was consumed?

Answer 1

fea O3 (s) 2Fels) +09 A 60 (a) 55.84591mel Fe = Molae mass mass Moles Fe = 55.805mol Molar mass 1.791 mele 159.69 atmol (C) Molar mass ferOs Meles fer 03 Mass 15Q.6 gfmel Molar Mass 0.626 mole dy msle fez o3 Fer 3 om 2 moles 81 Fe fosms meles From 5 moles fe 1x5 2.5 mole Le) 1.5 meles O oms 1 mole feos Or, moles Fubm #5 nmales feros 1 x5=3.33 misles Molarr mass x males Mars = maus fero3 =159. 69 2 X 50 moles mol fesos= 984-5 Maus a Mass Molar maus X males 1 6 /omet 160% X lo mel Mars 1 D