Question

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.


I initially thought it was A, but I don't know if there's enough calcium there for it to make a difference. So I'm torn now and I have no idea how to figure it out. V.V

Answer 1

@agent0smith got any ideas?

Answer 2

I think we'll first have to write the chemical reaction - you need to see if the initial reaction has an excess of water.

\[\Large Ca + H_2O \rightarrow H_2 + Ca O\]I think it looks balanced already.

Answer 3

Yeah, Ca ion is 2+ and O ion is 2-, so it's balanced. (hopefully this is making some sense)

Answer 4

\[\Large Ca + H_2O \rightarrow H_2 + Ca O\]since we only have 0.1 moles of Ca, we'll only need 0.1 moles of H2O right? Since from the equation, 1 mol Ca reacts with 1 mol H2O.

Answer 5

So we can just find out what the mass of 0.1 mol of H2O is...

1mol of H2O has an atomic mass of 18g (the 2 H's have a mass of 1 and the O is 16)

So 0.1mol of H2O will only have a mass of 1.8g - this is how much water is used when 0.1 mol Ca reacts with water.


Remember the original reaction has 880g of water. Then it's reduced to 440g.

Answer 6

So there was a definite excess of water and the decrease in water should make no difference?

Answer 7

Yep! there was like 490 times too much water, and now there's only 245 times too much water :P

Answer 8

Those numbers were estimates but I was pretty close! 488.889 times too much, then 244.44 times too much.

Answer 9

Thank you very much for the help :3

Answer 10

You're welcome!