Suppose 1.2x10^3 kg of C is mixed with 2.4x10^5g oxygen gas and carbon dioxide forms.
1. write the reaction and balance
2.which reactant is the limiting reagent and which reactant is in excess
3. How many grams of carbon dioxide will be formed.
I am wondering what you are having difficulty with.
a. write and balance the equation
b. note the mole ratio of C to O2
c. check the mole ratio of the amounts given.
d. if the given mole ratio is greater than b, then you have C in excess
To answer these questions, we need to follow a step-by-step process:
1. Write the reaction and balance it:
C + O₂ -> CO₂
In this reaction, carbon (C) reacts with oxygen gas (O₂) to form carbon dioxide (CO₂).
To balance the equation, we need an equal number of carbon atoms and oxygen atoms on both sides:
C + O₂ -> CO₂
2. Determine the limiting reagent and excess reagent:
To determine the limiting reagent, we compare the number of moles of each reactant and see which one is less. The reactant with fewer moles is the limiting reagent.
Given:
Mass of carbon (C) = 1.2x10^3 kg
Mass of oxygen gas (O₂) = 2.4x10^5 g
First, convert the mass of carbon (C) to grams:
1.2x10^3 kg = 1.2x10^6 g
Now, calculate the number of moles for each reactant:
Molar mass of C = 12 g/mol
Molar mass of O₂ = 32 g/mol
Moles of C = mass/molar mass = 1.2x10^6 g / 12 g/mol
Moles of O₂ = mass/molar mass = 2.4x10^5 g / 32 g/mol
Compare the moles to identify the limiting reagent:
Moles of C = 1.2x10^6 g / 12 g/mol = 1x10^5 mol
Moles of O₂ = 2.4x10^5 g / 32 g/mol = 7.5x10^3 mol
Since we have fewer moles of oxygen gas (O₂) compared to carbon (C), oxygen gas (O₂) is the limiting reagent. The carbon (C) is in excess.
3. Calculate the grams of carbon dioxide (CO₂) formed:
To do this, we need to use the stoichiometry of the balanced equation. From the balanced equation:
1 mole of C reacts to form 1 mole of CO₂.
Given that the moles of C is 1x10^5 from the previous calculations, we can determine the moles of CO₂ formed.
Moles of CO₂ = Moles of C = 1x10^5 mol
Now, calculate the mass of CO₂ using the molar mass of CO₂ (44 g/mol):
Mass of CO₂ = Moles of CO₂ x Molar mass of CO₂
Mass of CO₂ = 1x10^5 mol x 44 g/mol = 4.4x10^6 g
Therefore, 4.4x10^6 grams of carbon dioxide (CO₂) will be formed.