A chemist requires 0.719 mol Na2CO3 for a reaction. How many grams does this correspond to?

asked by @nicoles122 • about 1 year ago • Chemistry • 5 pts
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1 answer

Hey Nicole! So you just need to multiply the number of moles with the molar mass of the molecule.

First compute for the molar mass, you can get the atomic mass of each element from the periodic table. Also, the subscript for each molecule means you will have to multiply the atomic mass by that number:

AM Na = 22.99 g/mol AM C = 12.01 g/mol AM O =16.00 g/mol

MM Na2CO3 = 2(AM Na)+(AM C) + 3(AM O) = 2(22.99 ) + 12.01 + 3(16.00) = 105.99 g/mol

Now Get the mass

mass Na2CO3 = (moles Na2CO3)(MM Na2CO3) = (0.719 mol)(105.99 g/mol)

= 76.207 g

answered by @leot1 • about 1 year ago