SrH2 (s)+ 2H2O (l) >>> Sr(OH)2 (s) + 2H2 (g)

Question

Calculate the mass of hydrogen gas that can be prepared from 5.22 grams SrH2 and 4.35 grams of H2O.

I keep getting 0.729 grams but its wrong...

Answer 1

SrH2 + 2H2O ==> Sr(OH)2 + 2H2

I suspect this is a limiting reagent type problem.
Convert grams to moles.
moles SrH2 = g/molar mass = 5.22/89.64 = 0.0582
moles H2O = 4.35/18.015 = 0.241

Convert moles of each to moles of product.
moles H2 from SrH2 = 0.0582 x (2 moles H/1 mole SrH2) = 0.0582 x (2/1) = 0.116.

moles H2 from H2O = 0.241 x (2 moles H2/2 moles H2O) = 0.241 x (2/2) = 0.241 x 1 0.241

Both answer, of course, can't be correct (because they are different); the correct one is ALWAYS the smaller. In this case, moles H2 produced is 0.116 and SrH2 is the limiting reagent.

Then convert 0.116 moles H2 to grams.
g = moles x molar mass = 0.116 x 2.016 = 0.23386 which rounds to 0.234 to three significant figures.

I tried two or three ways to get your answer of 0.769 and couldn't so I don't know where you went wrong.
Check my arithmetic.

Answer 2

gracias amigo