Starting with the following equation,
Pb(NO₃)₂(aq) + K₃PO₄(aq) → Pb₃(PO₄)₂(s) + KNO₃(aq)
calculate the moles of Pb(NO₃)₂ that will be required to produce 495 grams of Pb₃(PO₄)₂.
5 answers
oof i'm glad i'm still in 8th grade
45 g of Pb₃(PO₄)₂ is 45/811.5 = 0.55 moles
Now we have to balance the equation
Pb(NO₃)₂ + K₃PO₄ → Pb₃(PO₄)₂ + KNO₃
First the lead:
3Pb(NO₃)₂ + K₃PO₄ → Pb₃(PO₄)₂ + KNO₃
Now the PO₄:
3Pb(NO₃)₂ + 2K₃PO₄ → Pb₃(PO₄)₂ + 6KNO₃
Now the NO₃ is also balanced
Now the equation tells you that each mole of Pb₃(PO₄)₂ requires 3 moles of Pb(NO₃)₂. So the answer to the question is
0.55 moles of Pb₃(PO₄)₂ requires 3*0.55 = 1.65 moles of Pb(NO₃)₂
Now we have to balance the equation
Pb(NO₃)₂ + K₃PO₄ → Pb₃(PO₄)₂ + KNO₃
First the lead:
3Pb(NO₃)₂ + K₃PO₄ → Pb₃(PO₄)₂ + KNO₃
Now the PO₄:
3Pb(NO₃)₂ + 2K₃PO₄ → Pb₃(PO₄)₂ + 6KNO₃
Now the NO₃ is also balanced
Now the equation tells you that each mole of Pb₃(PO₄)₂ requires 3 moles of Pb(NO₃)₂. So the answer to the question is
0.55 moles of Pb₃(PO₄)₂ requires 3*0.55 = 1.65 moles of Pb(NO₃)₂
Note that oobleck made a typo and the 45 g should be 495. Change the other numbers appropriately.
I actually slipped in two mistakes, using 0.55 moles instead of 0.055 moles.
and then, of course, the real typo -- missing the 495 grams.
But I'm sure that Anonymous, careful student that she is, noticed all that ...
Thanks for checking in, @Dr
and then, of course, the real typo -- missing the 495 grams.
But I'm sure that Anonymous, careful student that she is, noticed all that ...
Thanks for checking in, @Dr
Okay I know hes a killer and all but I just wanted to share this joke XD
Jeffery dhomar eating at 5 guys*
this dosn't taste like (5 guys) XD
Jeffery dhomar eating at 5 guys*
this dosn't taste like (5 guys) XD