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2 KI + Pb(NO3)2 - Phl, + 2 KNO,how many moles of lead(II) iodide (Pbly) are produced from 6.0 mol of potassium iodide (KI)?number of moles:molTOOLS*10%

2 KI + Pb(NO3)2 - Phl, + 2 KNO,how many moles of lead(II) iodide (Pbly) are produced-example-1
User BofA
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1 Answer

3 votes

Answer

3.0 mol

Step-by-step explanation

Given:

Moles of KI = 6.0 mol

Chemical reaction:

What to find:

The moles of lead iodide (PbI₂) produced.

Step-by-step solution:

From the given balanced chemical reaction above;

2 mol KI produced 1 mol PbI₂,

so 6.0 mol KI will produce:


\frac{6.0\text{ }mol\text{ }KI*1\text{ }mol\text{ }PbI_2}{2\text{ }mol\text{ }KI}=3.0\text{ }mol\text{ }PbI_2

The number of moles of PbI₂ produced = 3.0 mol

2 KI + Pb(NO3)2 - Phl, + 2 KNO,how many moles of lead(II) iodide (Pbly) are produced-example-1
User Dave L
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