Question

For the reaction 2HCl(aq)+Na₂CO₃(aq)→2NaCl(aq)+H₂O(l)+CO₂(g), calculate how many grams of each product form when 10.9 of HCl is used.

A) 213.8g NaCl, 2.0g H₂O, 4.4g CO₂
​B) 5.45g NaCl ,1.0g H₂O, 2.2g CO₂
C) 10.9g NaCl, 2.0g H₂O, 4.4g CO₂
D) 21.8g NaCl, 4.0g H₂O, 8.8g CO₂
​

Answer 1

When calculating the products formed from 10.9 g of HCl reacting with Na₂CO₃, stoichiometry provides a mass of 17.47 g for NaCl, 2.69 g for H₂O, and 6.58 g for CO₂. None of the provided answer choices match these calculated values, suggesting an error in the provided options.

Step-by-step explanation:

To calculate the mass of each product formed in the reaction between HCl and Na₂CO₃, we use stoichiometry. First, we must convert the given mass of HCl into moles, using its molar mass (M(HCl) = 36.46 g/mol). For 10.9 g HCl:

moles of HCl = 10.9 g / 36.46 g/mol = 0.299 moles

According to the balanced chemical equation, 2 moles of HCl react with 1 mole of Na₂CO₃ to produce 2 moles of NaCl, 1 mole of H₂O, and 1 mole of CO₂.

Therefore, for 0.299 moles of HCl (which is half the amount in moles):

• 0.299 moles of NaCl are produced (since there's a 1-to-1 mole ratio between HCl and NaCl in the products).

Mass of NaCl = moles of NaCl × molar mass of NaCl (M(NaCl) = 58.44 g/mol)

Mass of NaCl = 0.299 moles × 58.44 g/mol = 17.47 g

• 0.1495 moles of H₂O (since 2 moles of HCl produce 1 mole of H2O)

Mass of H₂O = moles of H₂O × molar mass of H₂O (M(H₂O) = 18.02 g/mol)

Mass of H₂O = 0.1495 moles × 18.02 g/mol = 2.69 g

• 0.1495 moles of CO₂ (due to the same 2:1 stoichiometric ratio)

Mass of CO₂ = moles of CO₂ × molar mass of CO₂ (M(CO₂) = 44.01 g/mol)

Mass of CO₂ = 0.1495 moles × 44.01 g/mol = 6.58 g

After calculations, none of the provided options (A, B, C, D) perfectly match the calculated masses of the products, indicating a potential error in the question or that the options don't align with the correct stoichiometry.