Given the equation: 2 Na (s) + H2CO3 (aq) ---> Na2CO3 (aq) + H2 (g)?

Question

Determine the theoretical yield of Na2CO3 if 9.54 grams of H2CO3 is reacted with 7.2 grams of Na. Identify the limiting and excess reactants

Answer 1

7.2/(2*22.98977) = 0.1565914
9.54/(2 + 12 + 48) = 0.1538710
The limiting reagent is H2CO3
The excess reagent is Na
The yields should be
Na2CO3
x/(46 + 12 + 48) = 9.54/62
x = 16.310 g

H2
x/2 = 9.54/62
x = 307.742 mg

Answer 2

First work out the moles
Moles = mass/MW

Moles Na = 7.2/ 23
=0.313

Moles H2CO3 = 9.54 / (2+12+3*16)
= 0.154

As you require 2 moles of Na for 1 mole of H2CO3 the limiting reagent will be the H2CO3. There will be an excess of Na.

As 1 mole H2CO3 reacts to form 1 mole Na2CO3, moles of Na2CO3 produced will be 0.154

Mass = Mole * MW
= 0.154 * (2*23+12+3*16)
= 16.32g

Note: For speed I have only used approx molecular weights (Ie from memory rather than looking them up). You need to go through the calcs with more accurate MW. There is a slim chance that the limiting and excess reactants may change as the mole ratios are close to stoichiometric.

Also make a note that there would be no Na left as any excess Na would react with the H2O from the aqueous H2CO3