Question

If 309 grams of potassium combines with 228 g chlorine gas, how many grams of the single product will be formed? What are the LR and the ER? Equation: 2 K + Cl2 -> 2 KCl

Answer 1

The amount of KCl that would be formed will be 478.8 grams

Stoichiometric calculations

From the equation of the reaction, the mole ratio of K to $Cl_2$ is 2:1.

Mole of 309 g potassium = 309/39 = 7.92 moles

Mole of 228 g Cl2 = 228/71 = 3.21 moles

Thus, potassium is in excess while Cl2 is limiting.

Mole ratio of Cl2 and KCl = 1:2

Equivalent mole of KCl = 3.21 x 2 = 6.42 moles

Mass of 6.42 moles KCl = 6.42 x 74.55 = 478.8 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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