The combustion of a sample of butane, c4h10, produced 2.46 grams of water.
2c4h10 + 13o2 -> 8co2 + 10h20

(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? and in grams?

(A) 0.14 moles water formed.

(B) 0.03 moles butane burned

(C) 5.7 grams of butane burned

(D) 0.18 moles of oxygen was used up in the reaction.

Explanation:

(A)

The given chemical reaction is as follows.

The number of water molecules in the reaction = 10

Therefore, 0.14 moles water formed.

(B)

Molarmass of butane = 180 g

The number of butane molecules in the reaction = 2

Therefore, 0.03 moles butane burned

(C)

Molar mass of oxygen = 32g/mol

The number of water molecules in the reaction = 10

The number of oxygen molecules in the reaction = 13

Therefore, 5.7 grams of butane burned

(D)

Molar mass of oxygen = 32 g/mol

Used amount of oxygen in the reaction = 5.7 g

Therefore, 0.18 moles of oxygen was used up in moles.

they are both synthetic elements

explanation:

d

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.