Just when we thought we were done with our conversions, here's a lesson with some more converting between moles and mass!
Something to remember though, that I think is handy with ANY of the conversions we learned: always look at the units, and think about what the final unit that you are trying to get is!
Something to remember though, that I think is handy with ANY of the conversions we learned: always look at the units, and think about what the final unit that you are trying to get is!
Some questions will give you an amount of moles and an equation (which will be balanced or you will have to balance it) and ask you to determine the mass of a compound/element
Converting moles to mass only requires one extra step from moles to moles, and its a step which we've actually learned before!
Converting moles to mass only requires one extra step from moles to moles, and its a step which we've actually learned before!
Here are some lovely examples:
Ex. How many grams of Bauxite (Al2O3) are required to produce 3.5 mol of pure Aluminum?
Step 1: Make an equation for this, and balance it!
2Al2O3 ---> 4Al + 3O2
Step 2: Look at the info you're given. You have the mol of Aluminum, and you want to find out the mass of the Bauxite. Find out the mol of Al2O3 (using the coefficients) so that you can convert it to mass (using molar mass!) :
3.5 mol Al x 2 mol Al2O3 = 1.75 mol Al2NO3 x 102 g = 178.5 g
4 mol Al 1 mol Al2NO3
Step 3: Sig Figs!
178.5 g = 1.8 x 10^2 g
178.5 g = 1.8 x 10^2 g
Ex. How many grams of water are produced if 0.84 mol of Phosphoric Acid is completely neutralized by Barium Hydroxide?
2H3PO4 + 3Ba(OH)2 ---> 6H2O + Ba2(PO4)2
0.84 mol H3PO4 x 6 mol H2O = 2.52 mol x 18.0 g H2O = 45 g
2H3PO4 + 3Ba(OH)2 ---> 6H2O + Ba2(PO4)2
0.84 mol H3PO4 x 6 mol H2O = 2.52 mol x 18.0 g H2O = 45 g
2 mol H3PO4 1 mol
Now here's a different kind of example. Instead of being given the moles of a compound, you are given the mass, and you must find the moles. You have to use the same conversions and all, just in a different order!
Ex. How many moles of Lead II Nitrate are consumed when 4.5 g of Sodium Sulphide completely reacts?
Pb(NO3)2 + Na2S ----> PbS + 2NaNO3
Ex. How many moles of Lead II Nitrate are consumed when 4.5 g of Sodium Sulphide completely reacts?
Pb(NO3)2 + Na2S ----> PbS + 2NaNO3
4.5 g Na2S x 1 mol = 0.0576 mol Na2S x 1 mol Pb(NO3)2 = 0.058 mol
78.1 g Na2S 1 mol Na2S
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