3AgNO3 + Na3PO4 ---> Ag3PO4 + 3NaNO3
Silver nitrate and sodium phosphate are reacted in equal amounts of 200. g each. How many grams of silver phosphate are produced?
1. What is the limiting?
2. How much silver phosphate is produced?
3. How much is in excess?
Ok this is what I got so far:
200gAgNO3= 1.18 mole AgNO3
200gNa3PO4= 1.22 mole NaPO4
Then 419g Ag3PO4 x3 = 1257Ag3PO4
limiting reagent- Ag3PO4?
figure out how many moles of each you have.
Then, multiply the moles of silver phosphate by three. Do you have that much or more of silver nitrate? If so, you have silver nitrate in excess, and the limiting reqeant is sodum phosphate. If the reverse is true, then silver nitrate is the limiting reqeant.
silver phosphate is produced this way, pick the limiting regeant. If it is sodum phosphat, you get the same number of moles of silver phosphate. If silver nitrate is the limiting reageant, then you get 1/3 of thos moles.
Hi bob!
Would you mind explaining this one too. Not sure I understand how to solve this one still. These are some of my first limited reagent questions I’ve ever done. Thanks!
do the steps I outlined, in order.
Moles silver nitrate
moles sodum phosphate.
Then
which is the moles silver phosphate greater than 3x moles silver nitrate?
limiting reageant?
then product will be ? same/or 3x?
I will be happy to critique your work.
Hi bob!
Would you mind explaining this one too. Not sure I understand how to solve this one still. These are some of my first limited reagent questions I’ve ever done. Thanks!
To find the answers to these questions, we need to use stoichiometry and the concept of limiting reactants.
1. To determine the limiting reactant, we can compare the moles of each reactant to the stoichiometry of the balanced equation. The balanced equation is:
3AgNO3 + Na3PO4 → Ag3PO4 + 3NaNO3
First, we need to convert the masses of AgNO3 and Na3PO4 to moles using their molar masses. AgNO3 has a molar mass of 169.87 g/mol, and Na3PO4 has a molar mass of 163.94 g/mol.
Moles of AgNO3 = 200 g / 169.87 g/mol
Moles of Na3PO4 = 200 g / 163.94 g/mol
Next, we compare the moles of both reactants to determine which one is the limiting reactant. The reactant that produces the least amount of product is the limiting reactant.
2. To find the number of grams of silver phosphate produced, we need to use the concept of stoichiometry. According to the balanced equation, 1 mole of Ag3PO4 is produced for every 3 moles of AgNO3 reacted. Therefore, the stoichiometric ratio is 1:3.
Moles of Ag3PO4 produced = Moles of limiting reactant (AgNO3) / Stoichiometric ratio (3 moles AgNO3: 1 mole Ag3PO4)
Finally, we convert moles of Ag3PO4 to grams using its molar mass, which is 418.57 g/mol.
Grams of Ag3PO4 produced = Moles of Ag3PO4 produced * Molar mass of Ag3PO4
3. To determine the excess reactant, we need to compare the moles of the limiting reactant with the moles of the other reactant. The reactant that has remaining moles after the reaction is completed is the excess reactant.
Moles of excess reactant = Moles of reactant in excess - Moles of limiting reactant
To calculate the grams of excess reactant, we multiply the moles of excess reactant by its molar mass.
Now, let's calculate the answers to these questions using the provided information.