Question

Given this reaction: 4K(s) + O2(g)→2K2O(s). Calculate how many grams of product is produced if 2.50 g of each reactant is reacted.
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Answer 1

Answer:

$m_{K_2O}^{by\ K} =3.01gK_2O$

Explanation:

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In this case, since 2.50 g of both potassium (molar mass = 39.1 g/mol) and gaseous oxygen (molar mass = 32.0 g/mol) react in a 4:1 and 1:2 mole ratio respectively, to produce potassium oxide (molar mass = 94.2 g/mol), we evaluate the mass of potassium oxide yielded by each reactant in order to identify the limiting one via stoichiometry:

$m_{K_2O}^{by\ K}=2.50gK*\frac{1molK}{39.1gK}*\frac{2molK_2O}{4molK}*\frac{94.2gK_2O}{1molK_2O} =3.01gK_2O\\\\m_{K_2O}^{by\ O_2}=2.50gO_2*\frac{1molO_2}{32.0gO_2}*\frac{2molK_2O}{1molO_2}*\frac{94.2gK_2O}{1molK_2O} =14.7gK_2O$

Thus, since the 2.50 g of potassium yields 3.01 g of potassium oxide, we infer it is the limiting reactant and that is the mass of produced product by the reaction.

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