Let's Calculate the Mass of Hydrogen Gas!
What is the balance formula for the reaction between SrH2 and H2O?
The balance formula is SrH2(s) + 2 H2O(l) => Sr(OH)2(s) + 2 H2(g).
How do we determine the limiting reactant in this reaction?
To determine the limiting reactant, we need to calculate the molar mass of SrH2 and H2O, then find the moles of each reactant used.
Calculation:
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
Since we have more moles of H2O, but for every mole of SrH2 used, it produces 4 moles of hydrogen atoms, the limiting reactant is SrH2.
Mass of hydrogen gas produced = 4 * 0.056450609 mol * 1.00794 g/mol = 0.227595307 g
Rounded to 3 significant figures, the mass of hydrogen gas produced is 0.228 g.
Isn't it exciting to see how the reactants in a chemical reaction interact to produce new substances? In this case, we explored the reaction between SrH2 and H2O, which results in the formation of Sr(OH)2 and hydrogen gas.
By calculating the molar masses of SrH2 and H2O, we determined that the limiting reactant in this reaction was SrH2. This means that all the SrH2 will be used up before all the H2O, resulting in the production of the maximum amount of hydrogen gas possible.
After performing the calculations, we found that 0.228 g of hydrogen gas can be produced from the given amounts of SrH2 and H2O. It's always fascinating to see how chemistry allows us to predict and understand the outcomes of reactions in such a precise manner.