Question

1. When 15.4 g of aluminum (26.98 g/mol) reacts with 14.8 g of oxygen gas (32.00 g/mol), what is the maximum mass (in grams) of aluminum oxide (101.96 g/mol) that can be produced? 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

2. If 14.1 moles of Cu and 48.7 moles of HNO₃ are allowed to react, how many moles of excess reactant will remain if the reaction goes to completion? Cu + 8 HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

3. Calculate the number of moles of excess reactant that will be left-over when 50.0 g of KI react with 50.0 g of Br₂: [ 2KI + Br₂   ->   2KBr + I₂ ]

Answer 1

1. From reaction

15.4 g Al *(1 mol Al/ 26.98 g)*(3 mol O₂ / 4 mol Al)*(32 g/ 1 mol) = 13.70 g O₂

Thus, Oxygen is a limiting reagent as amount of O₂ present is less

Now, Maximum mass of Al₂O₃ present;

14.8 g O₂ *( 1 mol/ 32 g )*(2 mol Al₂O₃ / 3 mol O₂)*(101.96 g/ 1 mol) = 31.4 g Al₂O₃

2. From reaction

14.1 mol Cu * ( 8 mol HNO₃ / 1 mol Cu) = 112.8 mol HNO₃

Thus, Cu in excess

48.7 mol HNO₃ * ( 1 mol Cu / 8 mol HNO₃) = 6.09 mol Cu

Excess mol of Cu = 14.8 - 6.09 = 8.71 mol of Cu

3. From reaction

50.0 g KI * ( 1 mol KI / 166 g)*( 1 mol Br₂ / 2 mol KI) * (159.8 g Br₂ / 1 mol) = 24.1 g Br₂

Thus, Br₂ is in excess and excess amount of Br₂ = 50.0 - 24.1 = 25.9 g / 159.8g/mol = 0.16 mol