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------------------------------------------------ In the reaction [tex]2K + Cl_2 \rightarrow 2KCl[/tex], how much potassium (K, 39.1 g/mol) is required to produce 298 g of KCl (Cl, 35.4 g/mol)?

Answer :

To produce 298 g of potassium chloride (KCl) in the reaction 2K + Cl2 → 2KCl, approximately 156.4 grams of potassium (K) is required.

To determine the amount of potassium (K) required to produce 298 g of potassium chloride (KCl), we need to use the stoichiometry of the balanced chemical equation.

The balanced equation is: 2K + Cl2 → 2KCl

From the equation, we can see that for every 2 moles of potassium (K), 2 moles of potassium chloride (KCl) are produced. Therefore, the mole ratio between K and KCl is 2:2 or 1:1.

To calculate the amount of potassium required, we can use the following steps:

1. Determine the molar mass of KCl:

Molar mass of KCl = (39.1 g/mol) + (35.4 g/mol) = 74.5 g/mol

2. Calculate the number of moles of KCl:

Moles of KCl = mass of KCl / molar mass of KCl

Moles of KCl = 298 g / 74.5 g/mol = 4 moles

3. Since the mole ratio between K and KCl is 1:1, the number of moles of potassium required is also 4 moles.

4. Calculate the mass of potassium required:

Mass of K = moles of K × molar mass of K

Mass of K = 4 moles × 39.1 g/mol = 156.4 g

Therefore, to produce 298 g of KCl, approximately 156.4 grams of potassium (K) is required.

To learn more about potassium Click Here: brainly.com/question/13321031

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