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.