What theoretical yield of NaCl would result from reacting 3.00 moles of Na2CO3 with excess HCl (aqueous)?

350.7 g of NaCl

Explanation:

For this reactions our reactants are:

HCl and Na₂CO₃.

Then, the reaction is:

2HCl + Na₂CO₃ →  2NaCl  +  H₂CO₃

2 moles of hydrochloric acid react to 1 mol of sodium carbonate in order to produce 2 moles of salt and 1 mol of carbonic acid.

Ratio is 1:2. 1 mol of sodium carbonate can produce 2 moles of NaCl

Then, 3 moles of carbonate may produce (3 . 2) /1 = 6

We convert the moles to mass to determine the theoretical yield

6 mol . 58.45g /mol = 350.7 g

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the answer is c. nacl