Stoichiometric calculation for cream neutralization:

• Acidity to reduce = $0.35\% - 0.15\% = 0.20\%\text{ LA}$.
• Mass of lactic acid to neutralize:
  $$M = 1200\text{ kg} \times \frac{0.20}{100} = 2.40\text{ kg}$$
• Neutralization reaction using sodium carbonate ($Na_2CO_3$, molecular weight $106.0\text{ g/mol}$):
  $$Na_2CO_3 + 2\text{CH}_3\text{CH(OH)COOH} \to 2\text{CH}_3\text{CH(OH)COONa} + H_2O + CO_2$$
  Since $1\text{ mol of } Na_2CO_3$ ($106\text{ g}$) neutralizes $2\text{ mol of lactic acid}$ ($2 \times 90.08 = 180.16\text{ g}$):
  $$\text{Neutralization factor} = \frac{106}{180.16} = 0.5884$$
• Mass of sodium carbonate required:
  $$\text{Mass} = 2.40\text{ kg} \times 0.5884 = 1.412\text{ kg}$$