Chemistry, 31.07.2019 20:00, loganharper992

If 5.10 g of acetic acid is reacted, how many grams of water can form?

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To determine how much water can form would depend on what the acetic acid is reacting with. However, a very common reaction of acetic acid that leads to the formation of water is the condensation that occurs during the esterification process. We can react acetic acid with an alcohol, such as methanol, and form the ester and water.

CH₃CO₂H + CH₃OH → CH₃CO₂CH₃ + H₂O

The mole ratio of water to acetic acid is 1:1. We can now use the mass of acetic acid to find the moles, which we can convert to moles of water:

5.10 g CH₃CO₂H / 60 g/mol = 0.085 mol CH₃CO₂H x 1 mol H₂O/1 mol CH₃CO₂H = 0.085 mol H₂O

0.085 mol H₂O x 18 g/mol = 1.53 g H₂O

In the reaction provided, assuming the reaction goes to completion, 5.10 g of acetic acid can form 1.53 g of water.

answered: Guest

out of these options:

ability to burn paper

pure sodium reacting violently with water

the boiling point of water

helium does not tend to react with anything

i would say this is a physical property:

helium does not tend to react with anything

why? :

i said helium does not tend to react with anything because well, it reacts to ! but ability to burn paper, pure sodium reacting violently with water, & the boiling point of water are chemical or

physical properties: the properties of a substance that can be observed without changing the chemical makeup of the substance

chemical properties: properties of a substance relating to the chemical nature and reactivity of the substance

answered: Guest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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