Given the following two step reaction:
4 P + 5 O2 -> P4O10
P4O10 + 6 H2O -> 4H3PO4
If you have 8.28 g of P, 15.65 g of O2, and 19.95 g of H2O, how many grams of H3PO4 can be produced?
4 P + 5 O2 -> P4O10
8.28g 15.65g
moles=g/mw x=15.65g/160.0
x=8.28/124.0 x= 0.0978
x=0.067
Which one would be limited? Please help if possible. Also, are my sigfigs correct?
4 P
8.28g
m=g/mw
x=8.28/124.0
x=0.067
+ 502
15.65g
x=15.65g/160
x= 0.0978
= P4O10
See the post above where I worked part of the problem.
40.9 grams produce
To find out how many grams of H3PO4 can be produced, we need to determine the limiting reactant among P, O2, and H2O.
1. Calculate the number of moles of each reactant:
- P: The molar mass of P is 31.0 g/mol. So, the number of moles of P = mass of P / molar mass of P = 8.28 g / 31.0 g/mol.
- O2: The molar mass of O2 is 32.0 g/mol. So, the number of moles of O2 = mass of O2 / molar mass of O2 = 15.65 g / 32.0 g/mol.
- H2O: The molar mass of H2O is 18.0 g/mol. So, the number of moles of H2O = mass of H2O / molar mass of H2O = 19.95 g / 18.0 g/mol.
2. Determine the limiting reactant:
To find the limiting reactant, we compare the mole ratios of the reactants and determine which reactant has the lowest number of moles.
From the balanced equation:
4 P + 5 O2 -> P4O10
P4O10 + 6 H2O -> 4H3PO4
The mole ratio between P and P4O10 is 1:1, and the mole ratio between P4O10 and H3PO4 is 1:4.
By using the law of conservation of mass, we compare the mole ratios of P to P4O10 and P4O10 to H3PO4.
We have: Moles of P4O10 = 4 * Moles of P
Moles of H3PO4 = 4 * Moles of P4O10
3. Calculate the moles of P4O10 produced:
Moles of P4O10 = Moles of P
4. Calculate the moles of H3PO4 produced:
Moles of H3PO4 = 4 * Moles of P4O10
5. Calculate the grams of H3PO4 produced:
Mass of H3PO4 = Moles of H3PO4 * Molar mass of H3PO4.
Remember to round the answer to an appropriate number of significant figures.
Now, using the given masses, follow the steps above to calculate the grams of H3PO4 that can be produced.