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EEJ MAIN Chemistry QUESTION #7526
Question 1

For the reaction $2\text{C}_{57}\text{H}_{110}\text{O}_6(s) + 163\ \text{O}_2(g) \to 114\ \text{CO}_2(g) + 110\ \text{H}_2\text{O}(l)$, find the mass of water produced from 445 g of C$_{57}$H$_{110}$O$_6$.

  • 490 g
  • 445 g
  • 495 g✔️
  • 890 g
Correct Answer Explanation

Molar mass of C$_{57}$H$_{110}$O$_6$: $57(12)+110(1)+6(16)=684+110+96=890\ \text{g mol}^{-1}$

Moles of C$_{57}$H$_{110}$O$_6 = \dfrac{445}{890} = 0.5\ \text{mol}$

From stoichiometry: 2 mol fat → 110 mol H$_2$O

So 0.5 mol fat → $\dfrac{110 \times 0.5}{2} = 27.5\ \text{mol H}_2\text{O}$

Mass of H$_2$O $= 27.5 \times 18 = \mathbf{495\ \text{g}}$