Final answer:
The volume of gaseous product after the reaction between 10 g of hydrogen and 64 g of oxygen is B) 2×22.4L, since two moles of water vapor are produced and one mole of gas at STP occupies 22.4 L.
Step-by-step explanation:
When 10 g of hydrogen and 64 g of oxygen are combined and exploded in a steel vessel, they react according to the balanced chemical equation:
2H₂(g) + O₂(g) → 2H₂O(g).
We know that 2 moles of hydrogen (2 x 2.02 g = 4.04 g) reacts with 1 mole of oxygen (32.0 g) to produce 2 moles of water (2 x 18.02 g = 36.04 g). Given the amounts of hydrogen and oxygen present, hydrogen is the excess reactant because only 32 g of hydrogen would be needed to react with 64 g of oxygen.
Following the reaction, for every 2 moles of gases (hydrogen and oxygen) that react, 2 moles of water vapor (gaseous product) are formed. From Avogadro's law, we know that 1 mole of gas at Standard Temperature and Pressure (STP) occupies 22.4 L. Hence, 2 moles of gas would occupy 2 x 22.4 L = 44.8 L.
This means the answer is B) 2×22.4L because the reaction produces two moles of water vapor.