You wish to calculate the mass of hydrogen gas that can be prepared from 5.06 g of srh2 and 4.34 g of h2o.

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

The angle of reflection is 30°