Question

What volume of carbon monoxide gas at STP is produced from 5.14 g calcium phosphate and 3.24 g silicon dioxide? (Use the equation provided.)

2Ca3 (PO4)2 + 6SiO2 + 10C → P4 + 6CaSiO3 + 10CO

Answer 1

haven't been talk this teacher is stupid

Answer 2

taught srry

Answer 3

So, first you have to identify the limiting reactant. Do you know how to do that?

Answer 4

nope i never seen any of this in my chem class

Answer 5

hm okay. do you know what "mole", "molar mass" are?

Answer 6

is it were its labeled g/mol?

Answer 7

that's the molar mass, yeah. But more importantly, do you know what a mole is?

Answer 8

nope?:/

Answer 9

it's an amount. Like a dozen is 12 items right?

A mole is \(6.022*10^{23}\) items. These can be molecules, atoms, ions, etc.

So, to the determine which if the reactants is limiting (meaning you can't make more because it's the least you have of) you first have to convert the masses (in grams typically) given to moles.

Do so by using: \(moles=\dfrac{mass}{Molar~mass}\)

Then you divide \(each\) reactants by it's stoichiometric coefficient in the balanced reaction (the number before the formula, for example for \(SiO_2\) it's 6).

Whichever reactant you have less of after dividing is the limiting reactant.

Answer 10

ill be right back

Answer 11

im srry I'm still kinda confused ? ok

Answer 12

sorry about that.

So, okay you have to convert the masses given:

5.14 g calcium phosphate
3.24 g silicon dioxide

to moles.

You need the molar mass of each of those reactants. Do you know how to get those?

Answer 13

so do u multiply 3.24 by the amount of mole in a gram?

Answer 14

nope, you divide the mass by the molar mass (for each reactant separately).

For \(Ca_3(PO_4)_2\), \(moles=\dfrac{5.14 ~g }{ 310~g/mol}=0.017 moles\)

Answer 15

ok

Answer 16

To find the limiting reactant, you also need to divide it by the stoichiometric coefficient in the balanced equation. For calcium phosphate is 2.

so 0.017/2= 0.008


Now, find the moles of Silicon dioxide.

Answer 17

ok

Answer 18

How many moles are there?

Answer 19

there are 0.0104 moles

Answer 20

did you divide 3.24 g by 60 g/mol, then by the coefficient 6?

Answer 21

nope opps math error

Answer 22

0.009

Answer 23

Okay, now that you know the moles present of each reactant, which is the limiting reactant?

Answer 24

umsilicon since its small ?

Answer 25

which one is smaller though?

Answer 26

silicon?

Answer 27

ok

Answer 28

oh

Answer 29

Okay, so now you have to make a ratio with the limiting reactant and what you're interested in.

We also have to use the original moles (0.017 moles) - before dividing.

Place the moles of each reactant and below it it's coefficient (form the balanced reaction).

\(\dfrac{moles~of~calcium ~phosphate}{6}=\dfrac{moles~of~CO}{10}\)


\(\dfrac{0.017~moles}{6}=\dfrac{moles~of~CO}{10}\)

Now, solve (algebraically) for moles of CO

Answer 30

ok

Answer 31

do u cross multiply?

Answer 32

@aaronq

Answer 33

yep

Answer 34

so its 0.17?

Answer 35

\(x=\dfrac{10*0.017}{6}\)

check you math

Answer 36

your*

Answer 37

oh i forget to divide 6? so its 0.283?

Answer 38

it's 0.0283, you missed a zero there

so thats the amount of moles of CO.

Now to convert it to volume, at STP, 1 mole of any gas = 22.414 L

so Volume of CO = 0.0283 moles *22.414 L/mol

Answer 39

ok so its 0.63?

Answer 40

yep, don't forget the units!

Answer 41

ok then??

Answer 42

@aaronq

Answer 43

and thats it.

The question was 'What volume of carbon monoxide gas at STP?'

that's what we found.

Answer 44

ur good can u help with two more questions?

Answer 45

yeah no problem. Do you mind opening a new question, this one is getting laggy.

Answer 46

ok no problem