Use the following balanced equation to answer the question below: 2c₇h₁₄+21o₂→14co₂+ 14h₂o

if 8 moles of c₇h₁₄ are added to 43 moles of o₂, how many grams of water are made?

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

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