Question

What is the maximum amount of P2O5 that can theoretically be made from 161 g of...

What is the maximum amount of P2O5 that can theoretically be made from 161 g of P4 and excess oxygen? I got 2.60 mol

What is the maximum amount of P2O5 that can theoretically be made from 176 g of O2 and excess phosphorus? 2.20 mol

Last is What is the percent yield if the actual yield from this reaction is 256 g ?

Homework Answers

Answer #1

The balanced equation is P4 + 5O2 2P2O5

Molar mass of P4 = 4xAt.mass of P = 4x31 = 124 g/mol

Molar mass of P2O5 = (2xAt.mass of P ) + ( 5xAt.mass of O )

                              = (2x31) + (5x16)

                              = 142 g/mol

According to the balanced equation ,

1 mole of P4 produces 2 moles of P2O5

                          OR

1x124 g of P4 produces 2x142 g of P2O5

161 g of P4 produces M g of P2O5

M = ( 161 x 2x142) / (1x124)

   = 368.7 g of P2O5

Therefore the maximum amount of P2O5 formed is 368.7 g

percent yield = ( actual yield / theoretical yield) x 100

                    = ( 256 / 368.7 ) x100

                    = 69.4 %

-----------------------------------------------------------------------------------------------

According to the balanced equation ,

1 mole of P4 produces 2 moles of P2O5

                          OR

1x124 g of P4 produces 2x142 g of P2O5

176 g of P4 produces M g of P2O5

M = ( 176 x 2x142) / (1x124)

   = 403.1 g of P2O5

Therefore the maximum amount of P2O5 formed is 403.1 g

percent yield = ( actual yield / theoretical yield) x 100

                    = ( 256 / 403.1 ) x100

                    = 63.5 %

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
P4+5O2→2P2O5 Consider a situation in which 112 g of P4 are exposed to 112 g of...
P4+5O2→2P2O5 Consider a situation in which 112 g of P4 are exposed to 112 g of O2. Part A What is the maximum amount in moles of P2O5 that can theoretically be made from 112 g of P4 and excess oxygen? Part B What is the maximum amount in moles of P2O5 that can theoretically be made from 112 g of O2 and excess phosphorus?
Consider a situation in which 235 g of P4 are exposed to 272 g of O2....
Consider a situation in which 235 g of P4 are exposed to 272 g of O2. What is the maximum number of moles of P2O5 that can theoretically be made from 235 g of P4 and excess oxygen? What is the maximum number of moles of P2O5 that can theoretically be made from 272 g of O2 and excess phosphorus? Express your answer to three significant figures and include the appropriate units.
Consider a situation in which 235 g of P4 are exposed to 272 g of O2....
Consider a situation in which 235 g of P4 are exposed to 272 g of O2. In Part A, you found the number of moles of product (3.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the number of moles of product (3.40 mol P2O5 ) formed from the given amount of oxygen and excess phosphorus. Now, determine the number of moles of P2O5 is produced from the given amounts...
1. Consider a situation in which 136 g of P4 are exposed to 144 g of...
1. Consider a situation in which 136 g of P4 are exposed to 144 g of O2. Part A What is the maximum amount in moles of P2O5 that can theoretically be made from 136 g of P4 and excess oxygen? Express your answer to three significant figures and include the appropriate units. Part B What is the maximum amount in moles of P2O5 that can theoretically be made from 144 g of O2 and excess phosphorus? Express your answer...
What is the maximum theoretical number of moles of aluminum oxide that can be recovered from...
What is the maximum theoretical number of moles of aluminum oxide that can be recovered from the reaction of 27.7 mol of aluminum and 27.8 moles of oxygen gas? 4 Al (s) + 3 O2(g) → 2 Al2O3(s) What mass of aluminum oxide can be recovered from the complete reaction of 34.3 g of Al with 34.3 g of oxygen gas? 4 Al (s) + 3 O2(g) → 2 Al2O3(s) If 8.7 moles of Al and 3.2 moles of Fe2O3...
Phosphorus and chlorine react as follows: P4(s) + Cl2(g) => PCl3(l). Balance it! Assume 135 g...
Phosphorus and chlorine react as follows: P4(s) + Cl2(g) => PCl3(l). Balance it! Assume 135 g P4(s) react with 333 g of Cl2(g). a) What is the limiting reactant? b) What is the reactant in excess and the amount in excess? c) What mass of phosphorus trichloride can be formed from the reaction?
Potassium permanganate, KMnO4, a common oxidizing agent, is made from various ores that contain manganese(IV) oxide,...
Potassium permanganate, KMnO4, a common oxidizing agent, is made from various ores that contain manganese(IV) oxide, MnO2. The following equation shows the net reaction for one process that forms potassium permanganate. a. 2MnO2 +2KOH+O2 → 2KMnO4 +H2 What is the maximum mass, in kilograms, of KMnO4 that can be made from the reaction of 453 g of MnO2 with 943 g of KOH and excess oxygen gas? b.Explain why the oxygen gas is in excess. c. If 1.18 kg of...
1. In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.8 g...
1. In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.8 g of lead(II) oxide. Calculate the percent yield of the reaction. 2Pb(s)+O2(g) = 2PbO(s) 2. Assuming an efficiency of 32.60%, calculate the actual yield of magnesium nitrate formed from 119.8 g of magnesium and excess copper(II) nitrate. Mg + Cu(NO3)2 = Mg(NO3)2 + Cu 3. Combining 0.282 mol of Fe2O3 with excess carbon produced 18.8 g of Fe. Fe2O3 + 3C = 2Fe + 3CO...
Using this reaction: 4Al(s) + 3O2(g) ----> 2Al2O3(s) a. Determine the limiting reactant and the maximum...
Using this reaction: 4Al(s) + 3O2(g) ----> 2Al2O3(s) a. Determine the limiting reactant and the maximum yield if 4.70g Al reacts with 5.00L of O2 at STP. ( I got Al as the limiting reactant and the maximum yield of .087mol.) b. If the actual yield is 8.03 grams, what is the % yield. ( I got 90.4%) Did I do these right?
The maximum amount of product D that can be produced from the following reaction is 55.75...
The maximum amount of product D that can be produced from the following reaction is 55.75 g. 2A + B → 2C + 3D Following an experiment based on this reaction, when the D was isolated, 50.67 g of D were obtained. What is the percent yield for this particular experiment?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT