What mass of hydrogen peroxide should result when 1.50 g of barium peroxide is treated?

Barium peroxide is BaO2.

Hydrogen peroxide is H2O2

This reaction produces hydrogen peroxide from barium peroxide:

BaO2 + H2SO4 ---> BaSO4 + H2O2

So, 1 mol of BaO2 produces 1 mol of H2O2.

Convert 1.50 g of BaO2 to moles:

moles = mass in grams / molar mass

molar mass of BaO2 = 137g/mol + 2*16g/mol = 169 g/mol

moles of BaO2 = 1.50g / 169 g/mol = 0.00888 moles

Then, the same number, 0.00888, will be produced of H2O2.

Now convert 0.00888 moles of H2O2 to mass in grams:

mass in grams = number of moles * molar mass

molar mass of H2O2 = 2 * 1 g/mol + 2 * 16 g/mol = 34 g/mol

mass of H2O2 = 0.00888 moles * 34 g/mol = 0.30192 grams.

Rounded to 3 significant figures = 0.302 grams.

0.302 grams

It is 4.92 e-4 sorry if i am wrong