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Question

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Explanation / Answer

V = 565 ml or 0.565 L of HCl

Na2CO3

forms m = 16.1 g of CO2 or mol = mass/MW = 16.1/44 = 0.366 mol of CO2

find HCl

4 HCl + 2 Na2Co3 --> 4 NaCl + 3 Co2 + 2 H2

find moles of HCl

if ratio is 4 mol of HCl per 3 mol of CO2, then ratio is 4/3

therefore

0.366 *4/3 = 0.4879 mol of HCl reacted

M = mol/V = 0.4879 / 0.565 = 0.8635 M of HCl