How many grams of water are produced when 37.7 g of hydrogen peroxide (H2O2) decomposes into oxygen gas and water?

Mass of water produced = 19.8 g

Explanation:

Given data:

Mass of hydrogen peroxide = 37.7 g

Mass of water produced = ?

Solution:

Chemical equation:

2H₂O₂      →        2H₂O  + O₂

Number of moles of H₂O₂:

Number of moles = mass/molar mass

Number of moles = 37.7 g/ 34 g/mol

Number of moles = 1.1 mol

Now we will compare the moles of H₂O₂ and H₂O from balance chemical equation in order to determine the moles of water formed.

                                   H₂O₂      ;           H₂O

                                      2         :             2

                                     1.1         :          1.1

Mass of water formed:

Mass = number of moles × molar mass

Mass = 1.1 mol × 18 g/mol

Mass = 19.8 g

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.