Question

What mass of water is produced by the combustion 1.2 x 10^26 molecules of butane ?

This is the last big project before my chemistry final and I am really struggling any help is appreciated:)

Answer 1

17.934 kg of water

Step-by-step explanation:

If balanced equation is not given; this format can come in handy.

For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:

2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O

For butane:

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)

2 moles of butane gives 10 moles of water.

1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)

Mass of 1 mole of any substance is equal to it's molar mass

So, if 2 x N molecules of butane gives 10 x 18 g of water.

Then 1.2 x 10²⁶ molecules will give:

`(1.2 * 10^(26) * 180)/(2 * 6.022 *  10^(23))`

= 17.934 x 10³ g of water

= 17.934 kg of water