a)
b) Combination or synthesis reactionc)
d) 30 g.
Explanation:
a. Balance the Equation
To balance the equation you must add the coefficients that yield the same number of atoms for the same kind of atoms on each side of the chemical equation.
You cannot modify the superscripts because the they are part of the chemical formula of the compound or molecule, they indicate the number of atoms of each kind inside the compound, i.e. they determine the composition substance.
The coefficients can be adjusted because they just indicate the number of units that react or are produced.
Starting from the skeleton equation:
Add an 8 before the SO₃ to balance the S atoms:
Now add a coefficient of 12 for the O₂ molecule on the left side:
If you count:
Atom Leff side Right side
S 8 8
O 24 24
Thus, the equation is balanced.
b. What type of chemical reaction is this?
When two elements or compounds combine to form one single compound that is a combination reaction.
The general form of a combination reaction is:
A + B → AB
This is called comibnation to distinguish from single replacement, double replacement, decompositon, and combustion reaction.
It is also named synthesis reaction because a new compound is being synthetized ("created") from simpler substances.
The combination of a non-metal and oxygen is a typical reaction of synthesis in which a basic oxide is produced. They are named basic oxides because they react with water to form a base: hydroxide.
c. What is the molar ratio of S₈, to O₂?
The mole ratio also called theoretical or stoichiometric mole ratio is the ratio of coefficients of the substances. The mole ratio of the reactants indicate the proportion in which the reactants combine to each other to form the product or products.
Then, take the coefficients of S₈ and O₂ from the balanced molecular chemical equation and form the ratio:
That means that every molecule (or mole of molecules) of S₈ reacts with 12 molecules (or mole of molecules) of O₂ to form the products: in this case 8 molecules (or mole of molecules of SO₃).
You use this ratio to determine the yield of the reactions in a procedure called stoichiometric calculations.
d. How many grams of O₂ should be used if 100 grams of SO₃, is the desired end product?
You will need the mole ratio of O₂ to SO₃ and the molar masses of both substanes.
Mole ratio:
Moles of SO₃
moles = mass in grams/molar massmolar mass of SO₃ = 80.066g/molmoles of SO₃ = 100g / 80.066g/mol = 1.25 mol SO₃
Moles of O₂
Conver to mass of O₂
molar mass of O₂ = 15.999 g/molmass of O₂ = 1.875mol × 15.999g/mol = 29.998 gRound to 1 significant figures: 30 g.
