I really need someone to explain this stoichiometry problem.
Directions:
Outline how to determine the limiting agent.
Determine the excess reactant and how much is left over.
Determine the amount of each of the products produced.
CaSO4 + AuF3 = CaF2 + Au2(SO4)3
Start with 15 grams of each reactant.
I know you have to balance the equation first. Then from there, I'm lost.
Please help me! :(
Here is a limiting reagent problem I've posted that gives exact detailed steps. These limiting reagent problems are a bit long and involved, especially when asking for the amount that remains unreacted (as this one does) but this procedure will work them.
Here is the link. Post your work if you get stuck and I shall be happy to help you through it. If you run into trouble be sure and tell me what you don't understand.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
Of course! I'll guide you through the steps to determine the limiting agent, excess reactant, and the amount of each product produced in this stoichiometry problem.
1. Balancing the equation:
First, balance the chemical equation by making sure the number of atoms on both sides is equal. The balanced equation for the reaction is:
CaSO4 + 3AuF3 = CaF2 + Au2(SO4)3
2. Determine the molar mass:
Next, determine the molar mass of each compound involved in the reaction. The molar mass represents the mass of one mole of a substance and is often expressed in grams per mole (g/mol). The molar mass for each compound in this equation is:
CaSO4 = 136.14 g/mol
AuF3 = 259.99 g/mol
CaF2 = 78.08 g/mol
Au2(SO4)3 = 685.98 g/mol
3. Convert grams to moles:
Now, convert the given mass of reactants to moles using their respective molar masses. We know that we start with 15 grams of each reactant, so:
Moles of CaSO4 = 15 g / 136.14 g/mol = 0.11 mol
Moles of AuF3 = 15 g / 259.99 g/mol = 0.06 mol
4. Determine the limiting agent:
To find the limiting agent, compare the mole ratio of the reactants from the balanced equation. The reactant that has a lower mole ratio is the limiting agent. In this case, for every CaSO4, we need 3AuF3. So, the limiting agent is AuF3 since we have an excess of CaSO4.
5. Determine the excess reactant and how much is left over:
To determine the excess reactant, subtract the amount used by the limiting agent from the initial amount. In this case:
Excess moles of CaSO4 = Moles of CaSO4 initially - Moles of CaSO4 used by the reaction
Excess moles of CaSO4 = 0.11 mol - 0.06 mol (since the reaction uses 3 times more AuF3 than CaSO4)
Excess moles of CaSO4 = 0.05 mol
To convert the excess moles of CaSO4 to grams, use its molar mass:
Excess grams of CaSO4 = Excess moles of CaSO4 * Molar mass of CaSO4
Excess grams of CaSO4 = 0.05 mol * 136.14 g/mol
Excess grams of CaSO4 = 6.81 g
Therefore, the excess reactant is CaSO4, and there will be 6.81 grams of it left over.
6. Determine the amount of each product produced:
To determine the amount of each product produced, use the mole ratios from the balanced equation. The mole ratio tells you how many moles of each product are formed from one mole of the limiting agent. From the balanced equation, we can see that:
1 mole of CaSO4 produces 1 mole of CaF2
1 mole of CaSO4 produces 1 mole of Au2(SO4)3
So, using the limiting agent (AuF3), we can determine the amount of each product as follows:
Moles of CaF2 = Moles of limiting agent (AuF3)
Moles of CaF2 = 0.06 mol
Moles of Au2(SO4)3 = Moles of limiting agent (AuF3)
Moles of Au2(SO4)3 = 0.06 mol
Finally, convert moles to grams using the molar masses of each product:
Mass of CaF2 = Moles of CaF2 * Molar mass of CaF2
Mass of CaF2 = 0.06 mol * 78.08 g/mol
Mass of CaF2 = 4.68 g
Mass of Au2(SO4)3 = Moles of Au2(SO4)3 * Molar mass of Au2(SO4)3
Mass of Au2(SO4)3 = 0.06 mol * 685.98 g/mol
Mass of Au2(SO4)3 = 41.16 g
Therefore, the amount of CaF2 produced is 4.68 grams, while the amount of Au2(SO4)3 produced is 41.16 grams.
Congratulations! You have successfully determined the limiting agent, excess reactant, and the amounts of each product produced in this stoichiometry problem.