Question

Calculate the mass of water produced when 5.87 g of butane reacts with excess oxygen.

Answer 1

Answer:

Explanation:

no. of moles of butane = 5.87g / 58.124 = 0.1009mol

since O2 is excess, limiting reagent is butane

from the equation, 2 moles of butane will reacted to produce water = 10 moles

therefore, 0.1009 moles of butane produced water = (0.1009 x 5) = 0.5050 moles

Weight of H2O formed = 0.5050 x 18.015 = 9.10 grams