Final answer:
The volume of hydrogen produced when 525 ml of 2.35 M HBr reacts can be calculated using stoichiometry and the molar volume at STP. The calculated volume is 13.818 L, which is not listed in the given options, indicating a potential error in the question or the options provided.
Step-by-step explanation:
To determine the volume of hydrogen gas collected when 525 ml of 2.35 M HBr undergoes a reaction, we should use the stoichiometry of the reaction and apply the ideal gas law. Since the reaction produces hydrogen gas at STP (standard temperature and pressure), we can use the molar volume of a gas at STP, which is 22.4 liters per mole.
First, calculate the moles of HBr using its concentration and volume. Moles of HBr = 2.35 M * 0.525 L = 1.23375 moles. The balanced equation for the reaction of HBr producing hydrogen is typically HBr -> H2 + Br2, which means 1 mole of HBr produces 0.5 mole of H2.
The moles of hydrogen gas produced will be half the moles of HBr, so moles of H2 = 1.23375 moles HBr * 0.5 = 0.616875 moles H2. Using the molar volume of a gas at STP, the volume of H2 = moles H2 * molar volume at STP = 0.616875 moles * 22.4 L/mol = 13.818 L.
Since the options provided in the original question do not contain this result, it seems there might be a mistake in the question or in the calculation. Please double-check the chemical equation and the reaction conditions provided. If the equation is as presented, with every mole of HBr producing half a mole of hydrogen, then the calculated volume would be larger than any of the options listed.