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The resulting product would be a roughly 2 M NaCl solution (slightly less than 2 molar because the solution is diluted by the water that is produced by the reaction).

2 M HCl + 1 M Na2CO3 --> 2 M NaCl + 1 M H2O + 1 M CO2

Two moles of NaCl weigh about 117 g (58.5 grams per mol) so the resulting solution has a sodium chloride concentration of about 117 grams per liter.

This concentration is well below the maximum solubility of water at room temperature and atmospheric pressure ( > 300 g/liter) so there would not be any NaCl at all precipitating.

The answer is thus: no NaCl precipitate will form.

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