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Mole Concept: Avoid Common Errors in Sec 3 Chemistry Calculations

Secondary 3 Chemistry students often face difficulties when tackling questions that involve both mole concept and chemical equations. A common stumbling block is understanding how limiting reagents affect the amount of precipitate formed, especially when solutions are mixed in varying volumes.

Take this question as an example:

When solutions of lead(II) nitrate and sodium iodide are mixed, lead(II) iodide is precipitated as shown by the equation:
Pb(NO₃)₂ (aq) + 2NaI (aq) → PbI₂ (s) + 2NaNO₃ (aq)

Same volumes of 1.00 mol/dm³ aqueous sodium iodide are added to different volumes of 0.500 mol/dm³ aqueous lead(II) nitrate in each of the four test tubes 1 to 4 as shown below.

Facing a tough Secondary 3 Chemistry question on mole concept and chemical calculation? Get on-demand homework help to support confident, independent learning.

Which statement is true after the precipitate settles?
A. The amount of precipitate in all four test tubes are the same.
B. The amount of precipitate increases consecutively from test tubes 1 to 4.
C. The amount of precipitate increases from test tube 1 to 2 but are the same in test tubes 3 and 4.
D. The amount of precipitate increases from test tubes 1 to 3, but are the same in test tubes 3 and 4.

Questions like this test conceptual understanding. Students who grasp the mole concept deeply can apply it across different scenarios, regardless of how the question is framed. Here’s how one student used instant, on-demand homework help to get guidance from a mentor and work through this step by step:

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Real mentors, Real-time answers

Ben

  • hello, i dont understand why the answer is C. However, i know that in test tube 1, NaI is limiting, in test tube 2 both reagents are just nice, while in test tubes 3 and 4 Pb(NO3)2 is in excess. however, how does that affect amount of ppt?
  • Hello!
  • hi
  • Let me take a look at the question
  • np
  • So you just need me to explain to you what’s happening in test tube 3 and 4?
  • yeah like how does lead (II) nitrate being excess or limiting affect the amt of ppt formed 
  • Since lead nitrate is in excess, technically we do not need to consider it.

    We just need to look at the number of moles of sodium iodide.

    One mole is sodium iodide produces one mole of lead iodide.
  • So for test tube 3 and 4, the number of sodium iodide is the same. Thus, the amount of lead iodide produces is the same for both test tubes.
  • Can you understand the explanation? ?
  • ohh ok so when lead nitrate goes from limiting to just nice ppt increases but when its in excess ppt remaisn the same? 
  • Yes, that’s right.

    For any equations, you do not need to consider the reactant that is in excess. You only consider the limiting reactant and do the calculation from there.
  • alright got it. thanks!

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