Answer:
∴grams of NaHCO₃(s) needed = 118.8 grams NaHCO₃
Step-by-step explanation:
HC₂H₃O₂ + NaHCO₃ => NaC₂H₃O₂(s) + H₂O(l) + CO₂(g)
(excess) (? grams) (112.0 grams = 112.0g/82.03g·mol⁻¹ = 1.37mol)
Since the reaction ratios are all 1:1 the moles of NaC₂H₃O₂(s) produced equals the number of moles of NaHCO₃(s) used.
Therefore, for 1.37 moles of NaC₂H₃O₂(s) produced one will need 1.37 moles of NaHCO₃(s).
Converting moles to grams, multiply by formula weight of NaHCO₃ (87.003 g/mole) ...
∴grams of NaHCO₃(s) needed = 1.37 mole x 87.003 g/mole = 118.8 grams NaHCO₃