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3 years ago

How many grams of oxygen are required to burn 13.5 g of acetylene

Chemistry

1 answer:

Vinil7 [7]3 years ago

8 0

Answer:

41.54 grams of oxygen are required to burn 13.5 g of acetylene

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₂H₂: 2 moles
O₂: 5 moles
CO₂: 4 moles
H₂O: 2 moles

Being the molar mass of the compounds:

C₂H₂: 26 g/mole
O₂: 32 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

C₂H₂: 2 moles* 26 g/mole= 52 grams
O₂: 5 moles* 32 g/mole= 160 grams
CO₂: 4 moles* 44 g/mole= 176 grams
H₂O: 2 moles* 18 g/mole= 36 grams

You can apply the following rule of three: if by stoichiometry 52 grams of acetylene react with 160 grams of oxygen, 13.5 grams of acetylene react with how much mass of oxygen?

$mass of oxygen=\frac{13.5 grams of acetylene*160 grams of oxygen}{52 grams of acetylene}$

mass of oxygen= 41.54 grams

41.54 grams of oxygen are required to burn 13.5 g of acetylene

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3 years ago

Read 2 more answers

Which of the following molecules has correctly labeled bond dipoles?

olga_2 [115]

Answer: The molecule identified with the letter C. HFO.

Explanation:

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3) O is more electronegative than H, so in the H-O bond the dipole is identified with the arrow from the H atom toward the O atom. That is correctly indicated in the drawing C, in this way:

-|----→

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4) F is more electronegative than O, so in the O-F bond the dipole is identified with an arrow from the O atom toward the F atom. That is correctly indicated in this way:

-|---→

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