How many moles of PbI2 are formed from this reaction? I keep getting the answer wrong. Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the following reaction: Pb(NO3)2 (aq) + 2NH4I(aq) → PbI2(s) + 2NH4NO3(aq) What volume of 0.390 M NH4I solution is required to react with 377 mL of a 0.180 M Pb(NO3)2 solution? How many moles of PbI2 are formed from this reaction? I have already figured out the answer to number 1, but I am struggling with number 2. I keep getting an answer of 0.06786 and, even after rounding, it is still marked incorrect. Please help.

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Anonymous

Nov 18, 2024

how many moles of PbI2 are formed from this reaction? i keep getting the answer wrong ??

How many moles of PbI2 are formed from this reaction? I keep getting the answer wrong.

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the following reaction:

Pb(NO3)2 (aq) + 2NH4I(aq) → PbI2(s) + 2NH4NO3(aq)

What volume of 0.390 M NH4I solution is required to react with 377 mL of a 0.180 M Pb(NO3)2 solution?
How many moles of PbI2 are formed from this reaction?

I have already figured out the answer to number 1, but I am struggling with number 2. I keep getting an answer of 0.06786 and, even after rounding, it is still marked incorrect. Please help.

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1 Answers

Anonymous

Oct 27, 2024

Unles you are rounding incorrectly (to 3 sig figs) you are correct and the given answer is wrong2.moles PbI = 0.06786 mol Pb(NO3)2 x (1 mol PbI2 / 1 mol Pb(NO3)2) = 0.06786 mol

= 0.0679 mol (3 sig figs)

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Anonymous

how many moles of PbI2 are formed from this reaction? i keep getting the answer wrong ??

1 Answers

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