Stoichiometry:
2.00 mol Pb(NO3)2 x 2 mol NaCl = 4.00 mol NaCl needed
1 mol Pb(NO3)2
3.00 mol NaCl x I mol PbCl x 242.65 g PbCl = 363.975 grams of PbCl
2 mol NaCl 1 mol PbCl
1.5 mol PbCl x 1 mol Pb(NO3)2 = 1.50 mol Pb(NO3)2 used up.
1 mol PbCl
2.00 mol available of Pb(NO3)2
- 1.50 mol used up of Pb(NO3)2
0.50 mol left over of Pb(NO3)2
Question: What is the volume of carbon dioxide gas at STP from the decomposition of 8.54g of lead carbonate?
PbCO3(s) > 2PbO(s) + CO2(g)
Stoichiometry:
8.54 g PbCO3 x 1 mol PbCO3 x 1 mol CO2 x 22.4 L CO2 = .7159 L CO2 at STP
267.21 g PbCO3 1 mol PbCO3 1 mol CO2
267.21 g PbCO3 1 mol PbCO3 1 mol CO2
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