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

17.94 kg of water

Explanation:

The combustion reaction of balanced butane (C₄H₁₀) is as follows:

2 C₄H₁₀ + 13 O₂ ⇒ 8 CO₂ + 10 H₂O

The amount of butane molecules can be transferred to moles to establish the stoichiometric relationship with water and calculate the moles of water that are formed:

6.02x10²³ molecules 1 mol of butane

1.2x10²⁶ molecules X = 199.34 moles of butane

Calculation: 1.2x10²⁶ molecules x 1 mol / 6.02x10²³ molecules = 199.34 moles of butane

According to the balanced equation:

2 moles butane 10 moles of water

199.34 moles of butane X = 996.68 moles of water

Calculation: 199.34 moles x 10 moles / 2 moles = 996.68 moles of water

And now, you can calculate the mass of the amount of moles obtained from water:

1 mole of water 18 g

996.68 moles of water X = 17.94 kg of water

Calculation: 996.68 moles x 18g / 1 mole = 17940.20g ≡ 17.94 kg of water

Therefore, the combustion of 1.2x10²⁶ butane molecules produces 17.94 kg of water.

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.