Limiting Reactant in Grams

Question

1.      What is the greatest amount of silicon tetrafluoride in grams that can be produced 48 grams of HF and 92 grams of SiO₂? Which reactant is limiting? How many grams of the excess reactant will be left after the reaction? SiO₂  + 4 HF ---> SiF₄ + 2 H₂O

 

2.      How many grams of sodium nitrate can be produced with 60 grams of sodium chloride and 55 grams of lead IV nitrate? Which reactant is limiting? How many grams of the excess reactant will be left after the reaction?

2 NaCl + Pb(NO₃)₂  ---> PbCl₂ + 2 NaNO₃

Answer 1

HI. I will help you set this problem up and explain how to find the limiting reactant for problem 1 and to calculate the excess, and I will let you solve problem 2 on your own because you will follow the same steps needed to solve problem 1.

Step 1: Before you do any calculations, check to make sure the reaction equation that you are provided is correctly balanced (ie make sure equal amounts of all atoms are present on both the products and reactants side), I will explain why this is necessary in the coming steps. The reaction you provided appears balanced, so we do not have to do anything for this step.

Step 2: Determine which reactant is the limiting reactant. The limiting reactant is going to be the substance that is completely consumed by the time the reaction is completed. In order to determine the limiting reactant, you must calculate how much of one reactant is needed to completely consume the other reactant. The easiest way to do this is to convert the masses of each reactant into moles using the molar masses of each reactant.

HF: 48 grams (1 mole / 20.01 grams) = 2.3988 moles of HF are available for the reaction

SiO₂ : 92 grams (1 mole / 60.08 grams) 1.5313 moles of SiO₂ are available for the reaction.

Now let's assume that all of our HF was consumed by the reaction. We would have to determine the amount of SiO2 needed to consume all of our HF using the coefficients from the balanced equation (this is why you need to balance your reaction equations before doing any calculations) :

2.3988 moles HF ( 1 mole SiO2 / 4 moles HF) = 0.5997 moles of SiO2 are needed to completely consume our 48 grams of HF

Now let's assume that all of our SiO2 was consumed by the reaction. We would determine this using the same steps above:

1.5313 moles SiO2 (4 moles HF / 1 mole SiO2) = 6.1252 moles of HF would be needed to consume all 92 grams of our SiO2.

Look back at the available amounts of each reactant (in moles). We only have 2.3988 moles of HF available, but consuming all of our SiO2 would require 6.1252 moles of HF (which we don't have). However, if we consumed all of our HF, it would only require 0.5997 moles of our SiO2 (which we do have because we have 1.5313 moles of SiO2 available). Since we have the ability to consume all of our HF, but we cannot consume all of our SiO2, we know that HF is our limiting reactant.

In order to calculate the excess reactant, you would subtract the moles of SiO2 that you need to consume all of the HF from the total moles of SiO2 that you have.

1.5313 moles available - 0.5997 moles needed = 0.9316 moles of SiO2 left over

You would then convert the excess moles to grams using the molar mass of SiO2 to determine the mass of the excess reactant.

0.9316 moles SiO2 (60.08 grams / 1 mole) = 55.97 grams of SiO2 left over.

I hope this helps! Please leave a comment if you need any further explanations or clarifications! Feedback is greatly appreciated! I am also available for lessons if you need extra help or practice with this.