Limiting Reactant And Yield, And Stoichiometry Help Please

I've had an A in my Chemistry class up till the last two chapters, but I'm starting to get very discouraged because I'm going for my degree in Laboratory Medical Science I need to know this stuff , but I keep getting stuck. Can someone please break down the process for me. I've read , and watched tutorials it's just not clicking. I got the balancing chemical equations down for the most part, but once I get passed that I'm stuck. Stoichiometry is just not sticking can someone . Here's an example of my homework I got the first two parts wrong. Can someone and work out the process I want the answers yes please, but please explain how you got it!
Part A
maximum amount of P2O5 that can theoretically be made from 211g P4 and excess oxygen?
Express your answer to three significant figures and include the appropriate units
Answer = 3.40 mol
Part B
maximum amount of P2O5 that can theoretically be made from 240g of O2 and excess phosphorus?
Express your answer to three significant figures and include the appropriate units.
Answer = 3.00 mol
Part C
In Part A, you found the amount of product (3.40mol~P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the amount of product (3.00mol~P2O5 ) formed from the given amount of oxygen and excess phosphorus.
Now, determine how much P2O5 is produced from the given amounts of phosphorus and oxygen.
Express your answer to three significant figures and include the appropriate units.
Answer = ?
Part D ? Don't know what this will be
Posted: May 10, 2013

Best Answer

You need to first write and balance the equation.
P4 + 5O2 = 2P2O5
If the equation isn't balanced, you aren't going to get the right answer.
Since the question says you have an excess of O2, the P4 is the limiting reactant and sets how much of the products you will make.
P4 = 123.90 g/mole so 211 g = 1.703 moles
From the equation, 1 mole of P4 makes 2 moles of P2O5 so you make 3.406 moles P2O5
Part B. Now the P4 is the excess. 240 g of O2 = 240g / 32 g/mole = 7.5 moles.
From the equation 5 moles of O2 makes 2 moles of P2O5. 7.5 moles O2 * 2 moles P2O5/5 moles O2 = 3.00 moles P2O5
Part C is a limiting reaction question.
Part A you had 1.703 moles of P4 and you had 7.5 moles of O2. To react with 1.703 moles P4, you need 5 times this amount of O2 or 8.51 moles. BUT, you only have 7.5 moles of O2 so O2 is the limiting reagent (you will have some P4 left over when all the O2 is used up).
Therefore the answer is the same answer you got in Part B, 3 moles of P2O5. Multiply by its molecular weight if you need g.

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