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asambeis [7]
3 years ago
14

24g of methane were burned in an excess of air. What mass of water would be produced in the reaction assuming complete combustio

n?Use the information below to answer the question.

Chemistry
1 answer:
expeople1 [14]3 years ago
8 0

Answer:

54g of water

Explanation:

Based on the reaction, 1 mole of methane produce 2 moles of water.

To solve this question we must find the molar mass of methane in order to find the moles of methane added. With the moles of methane and the chemical equation we can find the moles of water produced and its mass:

<em>Molar mass CH₄:</em>

1C = 12g/mol*1

4H = 1g/mol*4

12g/mol + 4g/mol = 16g/mol

<em>Moles methane: </em>

24g CH₄ * (1mol / 16g) = 1.5 moles methane

<em>Moles water:</em>

1.5moles CH₄ * (2mol H₂O / 1mol CH₄) = 3.0moles H₂O

<em>Molar mass water:</em>

2H = 1g/mol*2

1O = 16g/mol*1

2g/mol + 16g/mol = 18g/mol

<em>Mass water:</em>

3.0moles H₂O * (18g / mol) =

<h3>54g of water</h3>
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In a 66.0-g aqueous solution of methanol, CH 4 O , CH4O, the mole fraction of methanol is 0.300. 0.300. What is the mass of each
Mumz [18]

Answer:

The solution is composed by 37.5 g of water and 28.5 g of methanol.

Explanation:

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Therefore, mole fraction of water = 0.7

Let's find out the mass of each component by this two equations:

Methanol mass + Water mass = 66 g

Water mass = 66g - Methanol mass

Methanol mass = 66g - water mass

water mass / 18 g/mol = moles of water

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moles of water / total moles = 0.7

moles of methanol / total moles = 0.3

water mass / 18 g/mol / (water mass / 18 g/mol) + (methanol mass / 32 g/mol)  = 0.7 ; let's replace methanol mass, as (66 - water mass)

water mass / 18 g/mol / (water mass / 18 g/mol) + (66 - water mass / 32 g/mol)  = 0.7 → The unknown is water mass (X)

X / 18  / ( (X / 18)  + ((66-X) / 32)) = 0.7

(X / 18)  + ((66-X) / 32) = (16X + 594 - 9X)/288

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X / 18 = 7/2880 ( 7X + 594)

X = 7/2880 ( 7X + 594) 18

X = 7/160 (7X + 594)

X = 49/160X + 2079/80

X - 49/160X = 2079/80

111/160X = 2079/80

X = 2079/80 . 160/111 = 37.5 g → mass of water

Therefore, mass of methanol → 66 g - 37.4 g = 28.5 g

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