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How many moles of lead (II) chloride will be formed from a reaction between 6.5g of PbO and 3.2 g of HCl?
First of all, write out the reaction.PbO + 2HCl = PbCl2 + H2O(Don't forget to correctly balance the reaction)6.5g PbO x 1mol PbO/223 g = 0.03 mol PbO3.2g HCl x 1mol HCl/36.5g = 0.088 mol HClFrom this, we can conclude that PbO is the limiting reactant/reagent. Therefore, we will use PbO to figure out the amount of moles of PbCl2.0.03 mol PbO x 1mol PbCl2/1molPbO = 0.03 mol PbO.
First of all, write out the reaction.
PbO + 2HCl = PbCl2 + H2O(Don't forget to correctly balance the reaction)
6.5g PbO x 1mol PbO/223 g = 0.03 mol PbO
3.2g HCl x 1mol HCl/36.5g = 0.088 mol HCl
From this, we can conclude that PbO is the limiting reactant/reagent. Therefore, we will use PbO to figure out the amount of moles of PbCl2.
0.03 mol PbO x 1mol PbCl2/1molPbO = 0.03 mol PbO.
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