Question

if excess Na3 P04 reacts with 650.0cm^3 of 1.25M FeCl2 solution,what mass of Fe3(PO4)2 will be made

Answer 1

\[2Na _{3}PO _{4}+3FeCl _{2}\rightarrow Fe _{3}(PO _{4})_{2}+6NaCl\]

Answer 2

So 3 moles of FeCl2 is \[(3 \times 55.85) + (2 \times 35.45)=238.45g\]

Answer 3

would that be my answer?

Answer 4

No, it is part of my thought process. What you are looking for is this, you have a 3:1 ratio of FeCl2:Fe3(PO4)2, you are given a solution and need to figure out how many moles of it you have and then use the ratio to evaluate the amount they say you have.

Answer 5

So basically your limiting factor for this reaction is the FeCl2, now the part that I have difficulty with is the conversions from molarity to mass.

Answer 6

650 cm^3 is equal to 650ml.
\[650mL \times \frac{1.25moles}{1000 mL} = 0.8125 moles\]

Answer 7

\[\frac{3mole}{1mole}=\frac{0.8125mole}{xmol} \rightarrow 3x=0.8125 \rightarrow x=0.271mol\]

Answer 8

\[\frac{3mol FeCl_2}{6Fe_3(PO_4)_2}=\frac{0.8125mol FeCl_2}{x mol Fe_3(PO_4)_2} = 3x=4.875 \rightarrow x=1.625 mol Fe_3(PO_4)_2\]

Answer 9

(3*55.85)+(2(30.97+4*16.00)=357.49g

Answer 10

\[\frac{357.49g}{1mol} \times 1.625mol=580.9g\]

Answer 11

okay i think i get it thanks so much lol i have to go now im distracteed with te internet and i need to finish a worksheet

Answer 12

:)

Answer 13

thanks lol i was tourturing myself on this Q

Answer 14

You're welcome, it looks like it's right to me, but I would also suggest you compare it to your text. :)
Good luck.

Answer 15

thanks lol we dont use textbooks we use whatever the teacher gives us on notes thanks for the good luck too ima need it for my tests tomorrow