Question

Can anyone please help with this? 20 points

Answer 1

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