Question

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

Answer 1

Answer:

17.934 kg of water

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:

$\frac{1.2 \times 10^{26} \times 180}{2 \times 6.022 \times 10^{23}}$

= 17.934 x 10³ g of water

= 17.934 kg of water