Please make sure the answer is in the correct sigfigs!
Pb(CH3COO)2 + (NH4)2SO4 ----> 2NH4(CH3COO) + PbSO4
Given mass of lead(ii) acetate = 6.34 g
So, moles of lead(ii) acetate = 0.01949 mol
Number of moles of ammonium sulphate = 0.40 * 0.300 = 0.12 mol
Since 1 mol of lead(ii) acetate reacts with 1 mol of ammonium sulphate to give 2 mol of ammonium acetate and 1 mol lead sulphate.
Here limiting reagent is lead(ii) acetate.
So, moles of lead(ii) cation = moles of lead sulphate = moles of lead(II) acetate = 0.01949 mol
Volume of solution = 0.300 L
So, [Pb+2] = 0.065 M
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Please make sure the answer is in the correct sigfigs! Suppose 6.34 g of lead(II) acetate...
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