2H202--> 2H20 +O2

Given that you started with ½ cup of H2O2, what was the mass of each product in the reaction? (1/2 cup = 118.3 mL, density of H2O2 is 1.4 g/mL)

Convert volume of H2O2 (the 1/2 cup) to grams H2O2 using density.

Convert grams H2O2 to mols using mols = g/molar mass.
Convert mols H2O2 to mols H2O using the coefficients in the balanced equation.
Convert mols H2O to grams using mols x molar mass = grams.

Follow the same procedure for determining the grams of O2 in the product.

Post your work if get stuck.

To find the mass of each product in the reaction, we need to use the given information about the volume of H2O2 and its density. The given volume of H2O2 is 118.3 mL, and the density of H2O2 is 1.4 g/mL.

First, we need to convert the volume of H2O2 into its mass. We can do this by multiplying the volume by the density:

Mass of H2O2 = Volume of H2O2 * Density of H2O2
Mass of H2O2 = 118.3 mL * 1.4 g/mL
Mass of H2O2 = 165.62 g

Since the reaction is balanced, the molar ratio between H2O2 and the products (H2O and O2) is 1:1:1. This means that for every mole of H2O2, we will get one mole of H2O and one mole of O2.

To find the number of moles of H2O2, we can use its molar mass. The molar mass of H2O2 is 34.02 g/mol.

Number of moles of H2O2 = Mass of H2O2 / Molar mass of H2O2
Number of moles of H2O2 = 165.62 g / 34.02 g/mol
Number of moles of H2O2 = 4.862 mol

Since the molar ratio between H2O2 and the products is 1:1:1, we will get the same number of moles of each product.

Number of moles of each product = Number of moles of H2O2
Number of moles of each product = 4.862 mol

To find the mass of each product, we can multiply the number of moles by their respective molar masses.

Mass of H2O = Number of moles of H2O * Molar mass of H2O
Mass of H2O = 4.862 mol * 18.02 g/mol
Mass of H2O = 87.78 g

Mass of O2 = Number of moles of O2 * Molar mass of O2
Mass of O2 = 4.862 mol * 32.00 g/mol
Mass of O2 = 155.58 g

So, the mass of H2O produced in the reaction is 87.78 g, and the mass of O2 produced is 155.58 g.