HOC${}_6$H${}_4$COOH + CH${}_3$OH ⇌ HOC${}_6$H${}_4$COOCH${}_3$ + H${}_2$O
mHOC${}_6$H${}_4$COOH = 7600000.2,7.138/(152.80\%) = 23287500 gam ≈ 23,3 tấn