Zn +
HCI -
ZnCl2 +
H2

A. What is my theoretical yield of H2 if I start with 74 g of HCI?

B. If my percent yield was 93%, what was my actual yield?

1. Theoretical yield = 2.03g

2. Actual yield 1.89g

Explanation:

Let us write a balanced equation. This is illustrated below:

Zn + 2HCI —> ZnCl2 + H2

Molar Mass of HCl = 1 +35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 2 x 36.5 = 73g

Molar Mass of H2 = 2x1 = 2g/mol

1. From the equation,

73g of HCl produced 2g of H2.

Therefore, 74g of HCl will produce = (74 x 2)/73 = 2.03g

Therefore, theoretical yield = 2.03g

2. %yield = 93%

Theoretical yield = 2.03g

Actual yield =?

%yield = Actual yield /Theoretical yield x100

Actual yield = %yield x theoretical yield

Actual yield = 93% x 2.03 = (93/100)x2.03 = 1.89g

Actual yield =1.89g

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.

it is a solid

explanation: