Answer:
The answer to your question is: 516 g of water
Explanation:
2 C₇H₁₄ + 21 O₂ → 14 CO₂ + 14 H₂O
8 moles 43 moles
Process
1.- Find the limiting reactant
Ratio theoretical O₂ / C₇H₁₄ = 21 / 2 = 10.5
Ratio experimental O₂ / C₇H₁₄ = 43 / 8 = 5.3
As the ratio diminishes the limiting reactant is O₂.
2.- Calculate the moles of water
21 moles of O₂ ------------- 14 moles of water
43 moles of O₂ ------------ x
x = (43 x 14) / 21
x = 28.67 moles of water
3.- Calculate the grams of water
Molecular mass of water= 18g
18 g ------------------------ 1 mol of water
x ------------------------ 28.67 moles of water
x = (28.67 x 18) / 1
x = 516 g of water
