How many grams of potassium nitrate are required to produce 5.00 g of potassium nitrite according to the equation below?

2KNO3(s) ---> 2KNO2(s) + O2(g)

Use stoichiometry for this and the next question.

89.675

1.88 about 5.97 grams KNO3

To find out how many grams of potassium nitrate (KNO3) are required to produce 5.00 g of potassium nitrite (KNO2), we need to use stoichiometry.

First, let's start by writing the balanced chemical equation:
2KNO3(s) ---> 2KNO2(s) + O2(g)

The equation tells us that 2 moles of KNO3 react to produce 2 moles of KNO2. We need to convert grams to moles to determine the amount of KNO3.

1. Find the molar mass of KNO3:
K: 39.10 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (three oxygen atoms)

Adding up the atomic masses:
KNO3 = 39.10 + 14.01 + (16.00 x 3) = 101.10 g/mol

2. Convert the given mass of KNO2 to moles:
mass = moles × molar mass
5.00 g = moles × 101.10 g/mol

Solving for moles:
moles = 5.00 g / 101.10 g/mol ≈ 0.0494 mol

3. Use the stoichiometry of the balanced equation to relate KNO2 and KNO3:
From the balanced equation, we see that 2 moles of KNO3 are necessary to produce 2 moles of KNO2.

So, if 2 moles of KNO2 require 2 moles of KNO3, then 0.0494 mol of KNO2 will require:
(0.0494 mol KNO3 / 2 mol KNO2) × 2 moles KNO3 = 0.0494 mol KNO3

4. Convert moles of KNO3 back to grams:
grams = moles × molar mass
grams = 0.0494 mol × 101.10 g/mol ≈ 4.99 g

Therefore, approximately 4.99 grams of potassium nitrate (KNO3) are required to produce 5.00 grams of potassium nitrite (KNO2).