When 0.1 mol of calcium reacts with 880 g of water, 2.24 l of hydrogen gas form (at stp). how would the amount of hydrogen produced change if the volume of water was decreased to 440 ml (440 g)?

a: only half the volume of hydrogen would be produced.

b: the volume of hydrogen produced would be the same.

c: the volume of hydrogen produced would double.

d: no hydrogen would be produced.

Answer;

B: The volume of hydrogen produced would be the same.

Explanation;The reaction will be;

Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2 (g)

From the reaction; the mole ratio of calcium to water is 1 : 2; which means 1 mole of calcium requires 2 moles of water to produce 1 mole of hydrogen gas. Therefore; changing the mass of water used will not affect the amount of hydrogen gas produced since the amount of hydrogen gas produced depends on the amount of calcium used. In this case, water is in excess, because 0.1 moles of calcium requires only 0.2 moles of water which is 3.6 g.Hence; since the amount of calcium remains the same the amount of hydrogen gas produced will remain the same.