Question

The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water. 2 C4H10 + 13 O2 -------> 8 CO2 + 10 H2O (a) How many moles of water formed? (b) How many moles of butane burned? (c) How many grams of butane burned? (d) How much oxygen was used up in moles? In grams?

Answer 1

To determine the number of moles of water formed, divide the mass of water by the molar mass. The number of moles of butane burned can be calculated using the stoichiometry. The mass of butane burned can be calculated using the number of moles of butane and the molar mass.

Step-by-step explanation:

(a) To determine the number of moles of water formed, we need to use the balanced equation. From the equation, we can see that for every 10 moles of water produced, there are 2 moles of butane. Therefore, if 2.46 grams of water were formed, we can calculate the number of moles of water using the molar mass of water:

1. Number of moles of water = mass of water (g) / molar mass of water

(b) The number of moles of butane burned can be calculated using the stoichiometry of the balanced equation. From the equation, we can see that for every 2 moles of butane, 10 moles of water are produced:

1. Number of moles of butane = (number of moles of water formed) / (10/2)

(c) The mass of butane burned can be calculated using the number of moles of butane and the molar mass of butane:

1. Mass of butane burned = number of moles of butane * molar mass of butane

(d) The amount of oxygen used up in moles can be determined using the stoichiometry of the balanced equation. From the equation, we can see that for every 2 moles of butane, 13 moles of oxygen are used up:

1. Number of moles of oxygen used up = (number of moles of butane burned) * (13/2)

(e) The mass of oxygen used up in grams can be calculated using the number of moles of oxygen used up and the molar mass of oxygen:

1. Mass of oxygen used up = number of moles of oxygen used up * molar mass of oxygen

Answer 2

we are given with the amount of water produced in the reaction. In this case, we convert this into moles to conveniently find out other values of the compounds. In this case, a)0.1367 moles H20b)0.01367 moles butanec)0.7927 grams butaned)0.1776 moles O2 and 5.69 grams O2