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What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric acid (HCl)? Na2CO3 + 2HCl → 2NaCl + H2O + CO2

a) 8.50 g
b) 5.48 g
c) 10.88 g
d) 7.08 g

1 Answer

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Final answer:

The mass of sodium chloride formed when 7.5 g of sodium carbonate reacts with hydrochloric acid is calculated through stoichiometry to be 8.20 g, with the closest provided option being 8.50 g.

Step-by-step explanation:

To find the mass of sodium chloride (NaCl) formed from 7.5 g of sodium carbonate (Na2CO3), we first need to use stoichiometry. The balanced chemical equation for the reaction between sodium carbonate and hydrochloric acid (HCl) is:

Na2CO3 (aq) + 2HCl(aq) → 2 NaCl (aq) + CO2 (g) + H2O (l)

The molar mass of Na2CO3 is approximately 105.99 g/mol, and the molar mass of NaCl is approximately 58.44 g/mol. The stoichiometry of the reaction indicates that 1 mole of Na2CO3 produces 2 moles of NaCl. Therefore, we convert grams of Na2CO3 to moles, then use the stoichiometry to find moles of NaCl, and finally convert back to grams of NaCl:

7.5 g Na2CO3 × (1 mol Na2CO3 / 105.99 g Na2CO3) × (2 mol NaCl / 1 mol Na2CO3) × (58.44 g NaCl / 1 mol NaCl) = 8.20 g NaCl

Therefore, the correct answer is: 8.20 g which is not provided in the choices given but the closest option to the calculated mass is option a) 8.50 g.

User Blake Erickson
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