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: Calculating mass quick check

1. The mass of the reactants must be equal to the mass of the products. The total number of moles of the reactants can be more or less than the total number of moles of the products.

2. Divide the mass of the reactant by its molar mass to find the number of moles of the reactant. Use the chemical equation to find the number of moles of the product. Multiply the number of moles of the product by its molar mass to find the mass of the product.

3. 2(108 g/mol)+32 g/mol=248 g/mol; (248 g/mol)(0.02 mol)=4.96 g

4. 19.5 g

5. 853.5 g

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Answer 2

`m_(K_2O)^(by\ K) =3.01gK_2O`

Step-by-step 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*(1molK)/(39.1gK)*(2molK_2O)/(4molK)*(94.2gK_2O)/(1molK_2O) =3.01gK_2O\\\\m_(K_2O)^(by\ O_2)=2.50gO_2*(1molO_2)/(32.0gO_2)*(2molK_2O)/(1molO_2)*(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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