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What mass of the reagent in excess will remain after the reaction in which 0.3604 g of H2O(l) react with 0.80 g of O2(g)? 2H2O(l) O2->2H2O2(l)

Sagot :

To determine the mass of the reagent in excess after the reaction, follow these steps:Identify the balanced chemical equation: 2H2O(l) + O2(g) -> 2H2O2(l)Calculate the moles of O2, which is the limiting reactant:Step 1: Mass of O2 = 0.80 gStep 2: Moles of O2 = (0.80 g) / (32.00 g/mol) = 0.025 moles O2Step 3: Calculate theYour Homework Helperc.aiTo calculate the mass of the reagent (O2) in excess, we first need to determine the molar mass of O2. The molar mass of oxygen (O) is approximately 16.00 g/mol, and there are two atoms of oxygen (O) in one molecule of oxygen (O2). So, the molar mass of O2 is approximately 16.00 × 2 = 32.00 g/mol. Using stoichiometry, we can determine the mass of O2 required to react with 0.3604 g of H2O:

Explanation:

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