Hydrogen peroxide decomposes to form water and oxygen gas according to the following equation: 2H2O2(aq)  2H2O(l) + O2(g) If 315 g of hydrogen peroxide, H2O2, decomposes and all the O2 gas is collected in a balloon at 0.792 atm and 23C, what is the volume of the O2 gas collected?

141.89 dm^3

Explanation:

The equation of the reaction is;

2H2O2(aq) >2H2O(l) + O2(g)

Now , we are told that the mass of hydrogen peroxide decomposed was 315g. Number of moles of hydrogen peroxide in 315g of the substance is given by;

Number of moles= mass/molar mass

Molar mass of hydrogen peroxide= 34.0147 g/mol

Number of moles= 315g/34.0147 g/mol = 9.26 moles of hydrogen hydrogen peroxide.

From the reaction equation;

2 moles of hydrogen peroxide yields 1 mole of oxygen

9.26 moles of hydrogen peroxide yields 9.26 ×1/2 = 4.63 moles of oxygen

From the ideal gas equation;

Volume of the gas V= the unknown

Pressure of the gas P= 0.792 atm

Temperature of the gas= 23°C +273 = 296 K

Number of moles of oxygen = 4.63 moles of oxygen

R= 0.082atmdm^3K-1mol-1

Hence, from PV=nRT

V= nRT/P

V= 4.63 × 0.082 × 296/0.792 = 141.89 dm^3

the answer is d : ) have a good da