Question
24 mathrm(~g) of methane were burned in an excess of air. What mass of water would be produced in the reaction assuming complete combustion? Use the information below to answer the question. Enter your answer as a number g [ mathrm(CH)_(4)+2 mathrm(O)_(2) longrightarrow mathrm(CO)_(2)+2 mathrm(H)_(2) mathrm(O) ] Element & A, hydrogen & 1 carbon & 12 oxygen & 16
Answer
4.4
(33 Votes)
Bruce
Master · Tutor for 5 years
Answer
54 g
Explanation
To solve this problem, we first need to find out the molecular weight of Methane (CH4) and Water (H2O). 1. Methane (CH4) - Carbon (C) has a molar mass of approximately 12 grams/mole. - Hydrogen (H) has a molar mass of approximately 1 gram/mole. Therefore, - For Methane (CH4), its molecular mass = (12 * 1) + (1 * 4) = 16 grams/mole. 2. Water (H2O) - Oxygen (O) has a molar mass of approximately 16 grams/mole. - Therefore, the molar mass of water = (1 * 2) + (16 * 1) = 18 grams/mole Now, according to the provided chemical reaction, one mole of Methane (CH4) produces two moles of water H2O. Hence, if 16 g (1 mole) of CH4 will yield 36 g (18g*2moles) of H2O, we can derive the following relationship: Mass of methane : Mass of water = 16 g : 36 g We can use this ratio to determine the expected mass of H2O from 24 g of CH4: 24 g CH4 * (36g H2O / 16 g CH4) ≈ 54 g of H2O.