Question 580766
Chemistry questions (and even algebra questions) are likely to go unanswered here because there are too many questions. However, there is a website for chemistry questions where 99% or more of the questions get answered. It's
mychemistrytutor.com.
I used to answer questions there, but I was not really needed there. The need for help is much greater here.
The formula weight for {{{(NH[4])[2]CO[3]}}} can be calculated by combining the atomic masses of nitrogen (14.01), hydrogen (1.008), carbon (12.01) and oxygen (16.00).
{{{2(14.01+4*1.008)+12.01+3*16.00=2*18.042+12.01+48.00=96.094}}}
So we say that 1 mol of ammonium carbonate has a mass of 96.1 grams
{{{Pb[3](PO[4])[4]}}} would be the formula for lead IV phosphate
Its formula weight would be found by adding the proper combination of the atomic masses of lead (207.2), phosphorus (30.97) and oxygen (16.00)
{{{3*207.2+4(30.97+3*16.00)=621.6+4(30.97+64.00)=621.6+4*94.97=621.6+379.88=1001.48}}}
So we say that 1 mol of lead IV phosphate has a mass of 1001.5 grams.
{{{Pb(CO[3])[2]}}} would be the formula for lead IV carbonate.
We can calculate the formula weight as
{{{207.2+2(12.01+3*16.00)=207.2+2*(60.01)=207.2+120.02=327.22}}}
So we say that 1 mol of lead IV carbonate has a mass of 327.2 grams.
Looking at the formulas we figure that the balanced reaction must be
{{{Pb[3](PO[4])[4]+6(NH[4])[2]CO[3]}}}={{{3Pb(CO[3])[2]}}}+{{{4(NH[4])[3]PO[4]}}}
meaning that each mole of lead IV phosphate will react with 6 moles ammonium carbonate to yield 3 moles of lead IV carbonate.
So, the mass of ammonium carbonate needed is
(16.05 g Pb phosphate)(1 mol Pb phosphate/1001.5 g Pb phosphate)(6 mol NH4 carbonate/1 mol Pb phosphate)(96.1 g NH4 carbonate/1 mol NH4 carbonate = 9.24 g NH4 carbonate
{{{16.05*(1/1001.5)(6/1)(96.1/1)=9.24}}}
The mass of Pb IV carbonate produced would be
(16.05 g Pb phosphate)(1 mol Pb phosphate/1001.5 g Pb phosphate)(3 mol Pb carbonate/1 mol Pb phosphate)(327.2 g Pb carbonate/1 mol Pb carbonate = 15.73 g Pb carbonate
{{{16.05*(1/1001.5)(3/1)(327.2/1)=15.73}}}