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.
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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
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
To calculate the moles of Pb(NO₃)₂ required to produce 495 grams of Pb₃(PO₄)₂, we need to follow these steps:
Step 1: Determine the molar mass of Pb₃(PO₄)₂.
Pb₃(PO₄)₂ consists of 3 lead (Pb) atoms, 2 phosphate (PO₄) groups, and 8 oxygen (O) atoms in total.
Molar mass of lead (Pb): 207.2 g/mol
Molar mass of phosphate (PO₄): 94.97 g/mol
Molar mass of oxygen (O): 16.00 g/mol
Now, let's calculate the molar mass of Pb₃(PO₄)₂:
Molar mass = (3 * Pb) + (2 * PO₄) + (8 * O)
= (3 * 207.2) + (2 * 94.97) + (8 * 16.00)
= 621.6 + 189.94 + 128
= 939.54 g/mol
So, the molar mass of Pb₃(PO₄)₂ is 939.54 g/mol.
Step 2: Convert the given mass of Pb₃(PO₄)₂ to moles.
Mass = 495 grams
Molar mass = 939.54 g/mol
Moles = Mass / Molar mass
= 495 g / 939.54 g/mol
Calculating this gives us the number of moles of Pb₃(PO₄)₂.
Step 3: Use the balanced equation to find the moles of Pb(NO₃)₂ required.
From the balanced equation:
1 mole of Pb(NO₃)₂ produces 1 mole of Pb₃(PO₄)₂.
Therefore, the moles of Pb(NO₃)₂ required will be the same as the moles of Pb₃(PO₄)₂ calculated in step 2.
So, the moles of Pb(NO₃)₂ required to produce 495 grams of Pb₃(PO₄)₂ are equal to the moles calculated in step 2.