Question

What mass of carbon dioxide is formed when 64 KG of ethylene is burned in 142 KG of oxygen2c2h4+3o2=2co2+2h2o

Answer 1

mass of CO2 produced = 130 kg

Step-by-step explanation

Given:

`C_2H_4\text{ + 3O}_2\rightarrow\text{ 2CO}_2\text{ + 2H}_2O`

mass of C2H4 formed = 64 kg = 64000 g

mass of O2 = 142 kg = 142000 g

What we know:

Molar mass of C2H4 = 28,05 g/mol

Molar mass of O2 = 31.998 g/mol

Required: mass of CO2 produced.

Solution:

Step 1: Find the limiting reagent

C2H4:

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

n = 64000 g/28.05 g.mol

n = 2281.64 mol

divide by coefficient (from the equation)

n = 2281.64/1 = 2281.64 mol

O2:

n = 142000/31.998

n = 4437.8 mol

Divide by coefficient

n = 4437.8/3 = 1479.3 mol

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Therefore O2 is the limiting reactant, since it has less number of moles, meaning it will be consumed first in the reaction

Step 2: Use the moles of the limiting reagent to find the mass of CO2 produced.

Using the stoichiometry, the molar ratio between O2 and CO2 is 3:2

Therefore the number of moles of CO2 = 4437.8 mol x (2/3)

n of CO2 = 2958.5 mol

m = n x M

m = 2958.5 mol x 44,01 g/mol

m = 130205 g

m = 130 kg