Answer:
Mass of excess reactant left = 48 g
Explanation:
Given data:
Moles of carbon = 3.00 mol
Moles of oxygen = 3.00 mol
Mass of excess reactant left = ?
Solution:
Chemical equation:
2C + O₂ → 2CO
Now we will compare the moles of C and O₂ with CO to determine the excess reactant.
C : CO
2 : 2
3 : 3
O₂ : CO
1 : 2
3 : 2/1×3 = 6
Number of moles of CO produced by C are less thus it will act as limiting reactant. Oxygen is in excess.
Now we will compare the moles of oxygen and carbon to calculate the moles of oxygen used.
C : O₂
2 : 1
3 : 1/2×3 = 1.5 mol
Thus 1.5 moles of oxygen are used.
Number of moles of oxygen left = 3 - 1.5 = 1.5 mol
Mass of oxygen left:
Mass = number of moles × molar mass
Mass = 1.5 mol × 32 g/mol
Mass = 48 g
