Answer:
The mass of NaHCO₃ that the man would need to ingest to neutralize this much HCl is 0.059 g.
Step-by-step explanation:
1) Chemical equation (given):
- HCl (aq) + NaHCO₃ (aq) → NaCl (aq) + H₂O (l) + CO₂ (g)
Both sides have the same number of atoms of each element:
Element Let side Right side
H 1 + 1 = 2 1 × 2 = 2
Cl 1 1
Na 1 1
C 1 1
O 3 1 + 2 = 3
Hence, the equation is balanced.
2. Mole ratios:
- 1 mol HCl : 1 mol NaHCO₃ : 1 mol NaCl : 1 mol H₂O : 1 mol CO₂
2) Determine the number of moles of HCl in solution:
- M = n / V (liters) ⇒ n = M × V (liters)
- V (liters) = 200.mL (1 liter / 1000 mL) = 0.200 liter
- n = 0.035 M × 0.200 liter = 0.00070 mol of HCl
3) Set a proportion with the stoichiometric ratio and the actual ratio:
- 1 mol NaHCO₃ / 1 mol HCl = x / 00070 mol HCl
⇒ x = 0.00070 mol NaHCO₃
4. Convert 0.0070 mol NaHCO₃ to grams:
- Mass in grams = molecular mass × number of moles
- molecular mass NaHCO₃ = 84.007 g/mol
- Mass of NaHCO₃ = 84.007 g/mol × 0.00070 mol = 0.0588 g ≈ 0.059 g
The answer has two significant digits because the molarity (0.035 M) is reported with two signficant digits.