MgCl2 + Pb(NO3)2 = PbCl2 + Mg(NO3)2 Given 5 moles of magnesium chloride and 4 mo

Question

MgCl2 + Pb(NO3)2 = PbCl2 + Mg(NO3)2
Given 5 moles of magnesium chloride and 4 moles of lead (II) nitrate, calculate how many moles of lead (II) chloride can be made from each reactant. MgCl2 + Pb(NO3)2 = PbCl2 + Mg(NO3)2
Given 5 moles of magnesium chloride and 4 moles of lead (II) nitrate, calculate how many moles of lead (II) chloride can be made from each reactant. MgCl2 + Pb(NO3)2 = PbCl2 + Mg(NO3)2
Given 5 moles of magnesium chloride and 4 moles of lead (II) nitrate, calculate how many moles of lead (II) chloride can be made from each reactant. MgCl2 + Pb(NO3)2 = PbCl2 + Mg(NO3)2
Given 5 moles of magnesium chloride and 4 moles of lead (II) nitrate, calculate how many moles of lead (II) chloride can be made from each reactant.

Explanation / Answer

MgCl2 + Pb(NO3)2 -----------> PbCl2 + Mg(NO3)2

1 mol of MgCl2 reacts with 1 mol of lead nitrate, but we have 5 moles of magnesium chloride and 4 moles of lead nitrate. In this 1:1 relation, there's always a limitant reactant, and one in excess. In this case, magnesium chloride it's in excess and lead nitrate it's the limitant. So:

MgCl2 + Pb(NO3)2 = PbCl2 + Mg(NO3)2

i) 5 4 0 0

en) 1 0 4 4

So in conclusion, 4 moles of lead (II) chloride can be made.

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