From the balanced equation we can say that
2 mole of KI produces 1 mole of PbI2 so
6.0 mole of KI will produce
= 6.0 mole of KI *(1 mole of PbI2 / 2 mole of KI)
= 3.0 mole of PbI2
Therefore, the number of moles of PbI2 produced would be 3.0
For the chemical reaction 2 KI Pb(NO3)2 Pbl, + 2 KNΟ, how many moles of lead(II) iodide (Pbl2) are produced from 6....
For the chemical reaction 2KI+Pb(NO3)2⟶PbI2+2KNO3 how many moles of lead(II) iodide (PbI2) are produced from 8.0 mol of potassium iodide (KI)?
For the chemical reaction 2KI+Pb(NO3)2⟶PbI2+2KNO3 how many moles of lead(II) iodide (PbI2) are produced from 8.0 mol of potassium iodide (KI)?
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Assignment Soore: Give 1873/2800 Resources Question 18 of 28 > For the chemical reaction 2 KI+ Pb(NO,),2 Pbl,+2 KNO what mass of lead(II) iodide is produced from 4.23 mol of potassium iodide? g
Lead(II) nitrate and ammonium iodide react to form iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2NH4I (aq)-->Pbl2 (s)+2NH4NO3 (aq) ) What volume of Solution is required to react with 869 of 0.220 Pb(NO3)2 solution? How many moles of PbI2 are formed from this reaction. Question 7 of 17 ) Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO),(14) + 2NH,(aq) — PbL,(8) + 2NH, NO, (aq) What volume of a...
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Reaction 1: Use in question 3 Pb(NO3)2 (aq) + Kl (aq) → KNO, (aq) + Pblz (s) 3. a. When the reaction above is balanced how many moles of lead nitrate are required to react with 2.0 moles of potassium iodide? (1.0 mol Pb(NO3)2) b. How many grams of lead (II) iodide are produced from 5.0 moles of potassium iodide according to the equation given above? (1200 g Pblz)
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