Question

When coal is burned, One of the side products is carbon monoxide fill in the reaction

Answer 1

(a) m = 50.916 g

(b) CO2 is the limiting reagent and C is the reagent in excess.

(c) the mass excess left = 4.084 g

Step-by-step explanation:

Balance the given equation first:

`CO_2\text{ + C }\rightarrow2CO`

(a) Given:

mass of CO2 = 40.0 g

mass of C = 15.0 g

mass of CO = ?

To find the mass of CO that will be produced, we need to find the limiting reactant first. To find the limiting reactant we will calculate the number of moles of each reactant, the reactant with less number of moles is the limiting reactant.

CO2:

n = m/M where m is the mass and M is the molar mass

n = 40.0g/44.01 g/mol

n = 0.909 mol

C:

n = m/M

n = 15.0 g/12,0107 g/mol

n = 1.249 mol

CO2 is the limiting reagent and C is the reagent in excess.

Grams of CO that will be produced:

The molar ratio between CO2 and CO is 1:2

Therefore the number of moles of CO = 0.909 x 2 = 1.818 mol

m = n x M

m = 1.818 mol x 28,01 g/mol

m = 50.916 g

(c) To find how much of the reagent in excess will be left we will use the stoichiometry

n = 0.909 mol

m = 0.909 mol x 12.0107 g/mol

m = 10.916 g

15.0 g - 10.916 g = 4.084 g

Therefore the mass excess left = 4.084 g