Discussion: (24 points)

Butane, C4H10 burns in oxygen. How many grams of water vapor, H2O, are produced by the combustion of 580 grams of butane at stan

zepelin [54]

Answer:

The answer to your question is 900 g of water vapor

Explanation:

Data

mass of H₂O = ?

mass of butane = 580 g

Balanced chemical reaction

2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O

Process

1.- Calculate the molar weight of butane and water

Butane (C₄H₁₀) = 2[(12 x 4) + (1 x 10)]

= 2[48 + 10]

= 2[58]

= 116 g

Water (H₂O) = 10[(1 x 2) + (1 x 16)]

= 10[2 + 16]

= 10[18]

= 180 g

2.- Use proportions and cross multiplication to find the mass of water vapor

116 g of butane ------------- 180 g of water

580 g of butane ---------- x

x = (580 x 180) / 116

x = 900 g of water vapor

3 0

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Alona [7]

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5 0

4 years ago

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QUESTION 1

Fiesta28 [93]

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7 0

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identify the reagents necessary to achieve each of the following transformations o3 dms, pcc, ch2cl, h2so4,h2o,hgso4

hoa [83]

the reagents necessary to convert alcohol to ketone $Na_2Cr_2O_7 , H_2O$ which involves oxidation of alcohols.

<h3>What is oxidation of alcohols?</h3>

Alcohol oxidation is a significant organic chemistry process. Secondary alcohols can be oxidized to produce ketones, while primary alcohols can be oxidized to produce aldehydes and carboxylic acids.
In contrast, tertiary alcohols cannot be oxidized without the C-C bonds in the molecule being broken.
In order to cause primary alcohols to oxidize into aldehydes