Solving for the Molar Ratio of 2Na_{3}PO_{4}*12H_{2}0

[Winzer]

Homework Statement

Ok so I should know how to do this but I want to make sure.
The balanced equation involved is:
$$2Na_{3}PO_{4}*12H_{2}0+3BaCl_{2}*2H_{2}0\longrightarrow Ba_{3}(PO_{4})_{2}+6NaCl+30H_{2}0$$
In the lab I obtained 6.6e-4 moles of $$Ba_{3}(PO_{4})_{2}$$
and $$2Na_{3}PO_{4}*12H_{2}0$$ is limiting we found out.
I just need to convert moles of $$Ba_{3}(PO_{4})_{2}$$ to grams
$$2Na_{3}PO_{4}*12H_{2}0$$. I know how to convert from moles grams etc. but i can't decide when i do the mole to mole ratio. do I use 2*12=8moles for the $$2Na_{3}PO_{4}*12H_{2}0$$ or just 2?
Thanks

Homework Equations

$$2Na_{3}PO_{4}*12H_{2}0+3BaCl_{2}*2H_{2}0\longrightarrow Ba_{3}(PO_{4})_{2}+6NaCl+30H_{2}0$$

The Attempt at a Solution

I know how to convert from moles grams etc. but i can't decide when i do the mole to mole ratio. do I use 2*12=8moles for the $$2Na_{3}PO_{4}*12H_{2}0$$ or just 2?
Thanks

 

[symbolipoint]

1 mole of barium phosphate required 2 moles of sodium phosphate:12 hydrate. Straightforward stoichiometry. Arrange your whole conversion expression and include the units for each number. Start with the 1 mole of barium phosphate!

(1 mole ba.phos)*[(2 mole sod.phos.hydrate)/(1 mole ba.phos)]*(U grams/mole of sod.phos.hydrate)

You will need to write that all conventionally with pencil on paper; I used the symbol "U" for the number of grams per mole for sodium phosphate 12 hydrate, but you need to find it in a table or calculate it yourself.

 

[Winzer]

Yah know how to set it up, it just like i said before do I multiply with 24 or 2. Anyways I got 0.50g of $$2Na_{3}PO_{4}*12H_{2}0$$
and
0.48g of $$BaCl_{2}*2H_{2}0$$

 

[symbolipoint]

Yah know how to set it up, it just like i said before do I multiply with 24 or 2. Anyways I got 0.50g of $$2Na_{3}PO_{4}*12H_{2}0$$
and
0.48g of $$BaCl_{2}*2H_{2}0$$

You need to know what the numbers in the reaction statements mean. If you set up the calculation correctly, such as I have shown, then you will obtain the correct result. The two things applied in my setu-up expression which were not done were full-correct names for the compound (I used my own abbreviations) and the "U" in place of the actual formula weight.

 

[Winzer]

yah so:
$$(6.6e^-4 moles Ba_{3}(PO_{4})_{2}) *\frac{2BaCl_{2}*12H_{2}O}{1 mole Ba_{3}(PO_{4})_{2}}*\frac{380.12 2BaCl_{2}*12H_{2}O}{1 mole BaCl_{2}*12H_{2}O}$$

I get 0.50 right?

 

[Winzer]

right!?

 

[symbolipoint]

yah so:
$$(6.6e^-4 moles Ba_{3}(PO_{4})_{2}) *\frac{2BaCl_{2}*12H_{2}O}{1 mole Ba_{3}(PO_{4})_{2}}*\frac{380.12 2BaCl_{2}*12H_{2}O}{1 mole BaCl_{2}*12H_{2}O}$$

I get 0.50 right?

The only problem I see in the arranged expression is the formula weight is labeled as barium chloride - but you are not interested in that. You are interested in the sodium phosphate 12*hydrate. The the formula weight is the correct one for the phosphate.

 

[Weave]

Fix it to
$$(6.6e^-4 moles Ba_{3}(PO_{4})_{2}) *\frac{2Na_{3}PO_{4}*12H_{2}0}{1 mole Ba_{3}(PO_{4})_{2}}*\frac{380.12 2Na_{3}PO_{4}*12H_{2}0}{2Na_{3}PO_{4}*12H_{2}0}$$

 

[Winzer]

So is 0.50g right!?

 

[Winzer]

 

[symbolipoint]

YES. Probably closer to about 0.502 g. I hope this helps.
Why were you unsure?