Ask AI: in a lab, you reacted 3g of sodium carbonate with 0.003 mol of hydrochloric acid calculate the mass of the gas that should be produced

To calculate the mass of carbon dioxide (CO₂) produced, we need to first determine the limiting reactant in the reaction between sodium carbonate (Na₂CO₃) and hydrochloric acid (HCl). The chemical equation can be written as: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ Given that, you used 3 grams of Na₂CO₃ and 0.003 moles of HCl. Let's first calculate the number of moles of Na₂CO₃: Na₂CO₃ has two sodium (Na, with an atomic mass of 23 g/mol), one carbon (C, with an atomic mass of 12 g/mol), and three oxygen (O, with an atomic mass of 16 g/mol) atoms in its molecule. Therefore, its molar mass is: 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 g/mol. Moles of Na₂CO₃ = mass / molar mass = 3g / 106 g/mol = 0.0283 mol. Now, we compare the stoichiometry of Na₂CO₃ with HCl: 0.0283 mol Na₂CO₃ (2 mol HCl needed / 1 mol Na₂CO₃) = 0.0566 mol HCl needed. Since 0.0566 mol of HCl are needed and only 0.003 mol are available (provided), HCl is the limiting reactant. Thus, 0.003 moles of CO₂ will be produced. Finally, we calculate the mass of CO₂ produced: CO₂ has one carbon (C, with an atomic mass of 12 g/mol) and two oxygen (O, with an atomic mass of 16 g/mol) atoms making its molar mass: 12 + 2(16) = 12 + 32 = 44 g/mol. Mass of CO₂ produced = moles * molar mass = 0.003 mol × 44 g/mol = 0.132 g. Therefore, 0.132 g of carbon dioxide should be produced in this reaction.