(The reaction of calcium hydride and water produces calcium hydroxide and hydrogen as products). How many moles of H_2(g) will be formed in the reaction between 0.90mol CaH_2(s) and 1.33mol H_2O(l)?
This is a limiting reagent problem How do I know that? Because amounts for BOTH reactants are given.
Write and balance the equation.
Use the coefficients to convert mols of CaH2 to mols H2
Use the coefficients to convert mols H2O to mols H2.
The mols H2 obviously can't be two different numbers; the correct value in limiting reagent problems is AlWAYS the smaller value and the reactant producing that number is the limiting reagent.
To determine the number of moles of H2(g) formed in the reaction, we can first write the balanced chemical equation for the reaction:
CaH2(s) + 2H2O(l) -> Ca(OH)2(aq) + 2H2(g)
From the balanced equation, we can see that 1 mole of CaH2 reacts with 2 moles of H2O to produce 2 moles of H2.
Given that we have 0.90 mol of CaH2, we need to determine the limiting reactant. This is the reactant that will be completely consumed and will determine the amount of product formed.
To find the limiting reactant, we can use stoichiometry. We compare the moles of each reactant to their stoichiometric coefficients in the balanced equation.
Moles of H2O = 1.33 mol
Moles of CaH2 = 0.90 mol
From the balanced equation, we can see that 1 mole of CaH2 reacts with 2 moles of H2O. Therefore, using the ratio of CaH2 to H2O, we can calculate the moles of CaH2 that react with 1.33 mol of H2O:
0.90 mol CaH2 * (2 mol H2O / 1 mol CaH2) = 1.80 mol H2O
Since we have 1.33 mol of H2O, which is less than the calculated 1.80 mol H2O, it is clear that H2O is in excess and CaH2 is the limiting reactant.
Since 1 mole of CaH2 reacts to produce 2 moles of H2, the number of moles of H2 produced can be calculated as follows:
1 mole CaH2 * (2 mol H2 / 1 mole CaH2) = 2 moles of H2
Therefore, based on the limiting reactant, 0.90 mol of CaH2 will produce 2 moles of H2(g) in this reaction.