Question

1.For the reaction P4 O10(s) + 6H2O(l) → 4H3PO4(aq), what mass of P
must beconsumed if 5.51 × 10^23 molecules of H2O are also consumed?

can someone help me with this?

Answer 1

I'm just going to verify the chemical equation. Is this right?\[P_4O_10(s) + 6H_2O(l) \rightarrow 4H_3PO_4(aq)\]

Answer 2

yes :)

Answer 3

Ok, let's try this out. Do you know how to use dimensional analysis?

Answer 4

yes why?

Answer 5

Well, we're going to start with molecules of H2O:\[5.51\times 10^{23}\ molecules\ H_2O\]And we need to end at mass of P. Can you see what steps we need to take to get there?

Answer 6

that is my problem i know what to do but the formula that i did is was wrong

Answer 7

Ok, so I'll suggest you take this path with the dimensional analysis:

1) molecules of H2O to moles of H2O
2) moles of H2O to moles of P4O10
3) moles of P4O10 to moles of P4
4) moles of P4 to mass of P4
5) mass of P4 to mass of P

Make sense?

Answer 8

yes that is what i did but can i see your solution only in step 1?

Answer 9

Yes.\[5.51 \times 10^{23}\ molecules\ H_2O \times (\frac{1\ mol\ H_2O}{6.022\times 10^{23}\ molecules\ H_2O})=0.915mol\ H_2O\]

Answer 10

oooww i see now i know what i did wronng thank you :)

Answer 11

can i ask another question in here right now?

Answer 12

Sure!

Answer 13

Phosphorus, P, can be prepared from calcium phosphate by the reaction
3250 g 1665 g 690 g
2Ca_3(PO_4)2 4+ 6SiO_2 + 10C→ 6CaSiO +P_4 +10CO
310 g/mol 60.1 g/mol 12.0 g/mol 3

The molar mass for each reactant is shown below the reactant, and the mass of each
reactant for this problem is given above. Which reactant is the limiting reagent?

Answer 14

@Xishem

Answer 15

can you help me again @Xishem ?

Answer 16

Yes, just one second.

Answer 17

ok thanks :)

Answer 18

First convert each reactants mass to moles of that reactant.

Answer 19

then?

Answer 20

Then choose one of the reactants and convert it to moles of the other two. If the number you get for one of the other reactants is greater than the coefficient of that reactant, then the initial reactant you chose is NOT the LR.

For instance, if you get 10 moles of Ca3, you can convert that to the other two reactants:\[10mol\ Ca_3(PO_4)_2 \times (\frac{6mol\ SiO_2}{2mol\ Ca_3(PO_4)_3})=30mol\ SiO_2\]30 is greater than the coefficient of SiO2 in the chemical equation, which is 6, so that means that calcium phosphate is not the LR. The LR will produce results for all other reactants that are less than the coefficients in the chemical equation.

Does that make sense?

Answer 21

ohh thanks bro :) and earlier can you remember how many moles of P_4 there is?

Answer 22

thanke a lot men last one of the day because i have a test later on can you please write your entire solution about the first question @Xishem i really would appreciate it thanks :D

Answer 23

please @Xishem

Answer 24

1.For the reaction P4 O10(s) + 6H2O(l) → 4H3PO4(aq), what mass of P
must beconsumed if 5.51 × 10^23 molecules of H2O are also consumed?

This question?

Answer 25

yes bro :)

Answer 26

Sure. I'm going to do a written solution and upload it. Give me a few minutes.

Answer 27

ok thanks :)

Answer 28

thanks a lot men :)

Answer 29

No problem. Good luck!

Answer 30

uhhm w8 the answer is wrong the answer key says that it is 43.3g P_4 O_10

Answer 31

@Xishem

Answer 32

That is the mass of P4O10, not the mass of P. Just convert to mass of P4O10 instead of mass of P.

Answer 33

The problem you posted asks for the mass of P, which is why I solved for the mass of P.

Answer 34

THANKS! :D and sorry :)