If 2.5x10^-3 moles of O2 were produced in the reaction, how many grams of H2O2 had to react?
what reaction??
2H2O2 ==> 2H2O + O2
mols O2 = 2.5E-3
mols H2O2 = 2.5E-3 x (2 mol H2O2/1 molO2) = ?
Then grams H2O2 = mols x molar mass = ?
To find out how many grams of H2O2 had to react, we'll first need to determine the molar ratio between H2O2 and O2 in the reaction. From the balanced chemical equation of the reaction, we can identify that the stoichiometric coefficient of H2O2 is 2, while that of O2 is 1.
Next, we calculate the molar mass of H2O2, which is 34.0147 g/mol. This value can be found by adding up the atomic masses of hydrogen (1.00784 g/mol) and oxygen (15.999 g/mol) and multiplying them by 2 for the two hydrogen atoms.
Now, we can use the molar ratio and the amount of O2 (2.5x10^-3 moles) produced in the reaction to find the moles of H2O2 needed. Since the ratio is 1:2 (O2 to H2O2), we multiply the moles of O2 by the ratio (2 moles of H2O2 per 1 mole of O2).
(2.5x10^-3 moles O2) * (2 moles H2O2 / 1 mole O2) = 5x10^-3 moles H2O2
Finally, we multiply the mole value obtained by the molar mass of H2O2 to convert it to grams:
(5x10^-3 moles H2O2) * (34.0147 g/mol) = 0.1700735 grams H2O2
Therefore, approximately 0.170 grams of H2O2 had to react in the given reaction.