Identify the limiting and excess reagent for each of the following scenarios involving the formation of silicone dioxide, SiO2.
If you could just get geared in the right direction that would be wonderful. I am just a bit confused with this question. Thank you.
a. 2 atoms of silicone and 2 molecules of oxygen gas
b. 28.1 g of silicone and 48.0 g of oxygen gas
c. 6 mol of silicone and 2 mol of oxygen gas
First I want to correct a common error among many. Silicon is the element in chemistry. Silicone is, for want of a better word, a plastic. Silicones are wonderful stuffs. You can read more about them here.
http://en.wikipedia.org/wiki/Silicone
Now for your question.
The equation is
Si + O2 ==> SiO2
Look at the coefficients. The equation tells you that 1 mol Si will react with 1 mol oxygen to form 1 mol SiO2. To cut it down to size, if we have 2 atoms of Si, we will need 1 molecules of O2 to react with 1 atom Si and another molecule of O2 to react with the other Si so if have 2 molecules of O2, everything comes out even and neither Si nor O2 is in excess.
b. Put this in moles.
28.1g/atomic mass Si = 1 mol
48.0g O2/32 = 1.50 moles.
Since they react 1 mol to 1 mol, Si is the limiting reagent and we have some O2 remain unreacted when all is done.
c. I will leave this for you.
To identify the limiting and excess reagent in each scenario, you need to compare the stoichiometry of the balanced chemical equation with the given amounts of reactants.
First, start by writing the balanced chemical equation for the formation of silicon dioxide, SiO2:
Si + O2 → SiO2
a. In this scenario, you have 2 atoms of silicon and 2 molecules of oxygen gas. To determine the limiting and excess reagent, you need to convert these quantities to moles. Recognize that 1 molecule of oxygen gas (O2) contains 2 moles of oxygen (O), so you have:
2 atoms of silicon = 2/1 = 2 moles of silicon
2 molecules of oxygen gas = 2 x 2/1 = 4 moles of oxygen
To determine the limiting reagent, you need to compare the stoichiometric ratio of the balanced equation. The ratio is 1:1 for silicon to oxygen. Since you have 2 moles of silicon and 4 moles of oxygen, silicon is the limiting reagent because it will require only 2 moles of oxygen to react completely. Oxygen is in excess in this scenario.
b. Now, let's move to the next scenario where you have 28.1 g of silicon and 48.0 g of oxygen gas. To identify the limiting and excess reagents, convert these masses to moles using the molar mass of each element.
Molar mass of silicon (Si) = 28.1 g/mol
Molar mass of oxygen (O2) = 32.0 g/mol (16.0 g/mol per oxygen atom)
Moles of silicon = 28.1 g / 28.1 g/mol = 1.0 mol
Moles of oxygen = 48.0 g / 32.0 g/mol = 1.5 mol
Again, compare the stoichiometric ratio of the balanced equation, which is 1:1 for silicon to oxygen. In this case, silicon is the limiting reagent since it requires only 1 mole of oxygen to react completely. Oxygen is in excess.
c. In the last scenario, you have 6 mol of silicon and 2 mol of oxygen gas. Since these quantities are already given in moles, you can directly compare them to identify the limiting and excess reagents.
Stoichiometric ratio of Si to O2 is 1:1. Comparing the given quantities, we see that oxygen (2 mol) is the limiting reagent because it will require only 2 mol of silicon to react completely. Silicon is in excess.
Remember, the limiting reagent is the reactant that is completely consumed in the reaction, while the excess reagent is the reactant that is left over after the reaction is complete.