To make a phosphorus fertilizer, agricultural companies use the following reaction:
Ca3P2O8 + 2H2SO4 --> CaH4P2O8 + 2CaH4SO6
(fertilizer)
If 1.50 * 10^4 grams of H2SO4 are reacted with excess Ca3P2O8 and H2O, how many grams of fertilizer can be made?
Hi, I pretty much get what the problem is asking, but I'm getting confused on what to do with the two excess compounds,Ca3P2O8 and H2O, because I've only worked on problems with one compound. It would be great if someone could explain what I should do. -- Thank you.
You just ignore the excess compounds.
all that is important here is for every 2 mols of H2SO4 you get one mol of CaH4P2O8
MY FEED BACK PLEASE
To solve this problem, you need to determine the limiting reactant, which is the reactant that gets completely consumed first and determines the amount of product that can be formed.
In this reaction, you have excess Ca3P2O8 and H2O. These excess reactants will not get consumed completely and will not affect the amount of product formed.
To find the limiting reactant, you need to compare the moles of H2SO4 with the moles of Ca3P2O8.
1. Calculate the moles of H2SO4:
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
Given: mass of H2SO4 = 1.50 * 10^4 grams
molar mass of H2SO4 = 98.09 g/mol
moles of H2SO4 = (1.50 * 10^4) / 98.09
2. Now, use the balanced equation to determine the mole ratio between H2SO4 and Ca3P2O8:
From the balanced equation: 1 mol Ca3P2O8 reacts with 2 mol H2SO4.
moles of Ca3P2O8 = moles of H2SO4 * (1 mol Ca3P2O8 / 2 mol H2SO4)
3. Calculate the mass of fertilizer that can be formed:
Use the mole ratio between Ca3P2O8 and fertilizer from the balanced equation:
From the balanced equation: 1 mol Ca3P2O8 produces 1 mol fertilizer.
mass of fertilizer = moles of Ca3P2O8 * molar mass of fertilizer
Now you can solve the problem step-by-step using the above calculations.
Hello! I can help you with this problem.
In the given reaction, we are provided with the amount of H2SO4 (1.50 * 10^4 grams) and we need to determine the amount of fertilizer that can be made.
To do this, we need to identify which compound is the limiting reactant and calculate the amount of product formed based on the stoichiometry of the reaction.
The limiting reactant is the reactant that is completely consumed in the reaction, thereby limiting the amount of product that can be formed.
To find the limiting reactant, we can compare the moles of H2SO4 and Ca3P2O8 and see which one produces the lesser amount of product.
First, we need to calculate the moles of H2SO4 using its molar mass. The molar mass of H2SO4 is:
1 mol H2SO4 = 1 * (1.008 g/mol) + 4 * (16.00 g/mol) + 32.06 g/mol = 98.09 g/mol
Number of moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
= (1.50 * 10^4 g) / (98.09 g/mol)
= 152.8 mol
Now, let's calculate the moles of Ca3P2O8. Since Ca3P2O8 is in excess (it is mentioned as excess in the problem), we don't need to calculate its moles. This means that all of the Ca3P2O8 will react and there won't be any left over.
Next, we need to find the stoichiometry of the reaction, which tells us the mole ratio between H2SO4 and fertilizer (CaH4P2O8).
From the balanced equation:
1 mole of Ca3P2O8 + 2 moles of H2SO4 -> 1 mole of CaH4P2O8
Now, let's calculate the moles of fertilizer (CaH4P2O8) produced using the mole ratio:
Number of moles of CaH4P2O8 = 2 * moles of H2SO4
= 2 * 152.8 mol
= 305.6 mol
Finally, to calculate the mass of the fertilizer (CaH4P2O8) produced, we can multiply the number of moles by its molar mass. The molar mass of CaH4P2O8 can be calculated by summing up the individual atomic masses:
1 mol CaH4P2O8 = (1 * 40.08 g/mol) + (4 * 1.008 g/mol) + (2 * 31.00 g/mol) + (8 * 16.00 g/mol) = 188.22 g/mol
Mass of CaH4P2O8 = number of moles of CaH4P2O8 * molar mass of CaH4P2O8
= 305.6 mol * 188.22 g/mol
= 57,494.912 g
≈ 57,495 g (rounded to 4 significant figures)
Therefore, approximately 57,495 grams of fertilizer (CaH4P2O8) can be made when 1.50 * 10^4 grams of H2SO4 are reacted with excess Ca3P2O8 and H2O.
I hope this explanation helps! Let me know if you have any further questions.