<span>0.228 g
The balance formula for the reaction between SrH2 and H2O is
SrH2(s) + 2 H2O(l) ==> Sr(OH)2(s) + 2 H2(g)
So for every mole of SrH2 used, 2 moles of hydrogen gas, or 4 moles of hydrogen atoms are released. So let's calculate the molar mass of SrH2 and H2O so see what the limiting reactant is.
strontium = 87.62
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of SrH2 = 87.62 + 2 * 1.00794 = 89.63588 g/mol
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol
Moles of SrH2 = 5.06 g / 89.63588 g/mol = 0.056450609 mol
Moles of H2O = 4.34 g / 18.01488 g/mol = 0.240911957 mol
Looking at the balanced formula, for every mole of SrH2, it takes 2 moles of H2O. So the limiting reactant will be the SrH2. And for every mole of SrH2 used, we get 4 moles of hydrogen atoms. So
4 * 0.056450609 mole * 1.00794 g/mole = 0.227595307 g
Since we only have 3 significant figures, round the result to 3 figures, giving
0.228 g</span>
The answer is potential energy...
Hope this helped :)
