We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
It's just H20 but with 3 water molecules
Answer:
The volume of CO2 produced is 6.0 L (option D)
Explanation:
Step 1: Data given
Volume of oxygen = 3.0 L
Carbon monoxide = CO = in excess
Step 2: The balanced equation
2 CO (g) + O2 (g) → 2 CO2 (g)
Step 3: Calculate moles of O2
1 mol of gas at STP = 22.4 L
3.0 L = 0.134 moles
Step 3: Calculate moles of CO2
For 2 moles CO we need 1 mol of O2 to produce 2 moles of CO2
For 0.134 moles O2 we'll have 2*0.134 = 0.268 moles CO2
Step 4: Calculate volume of CO2
1 mol = 22.4 L
0.268 mol = 22.4 * 0.268 = 6.0 L
The volume of CO2 produced is 6.0 L