you wish to calculate the mass of hydrogen gas that can be prepared from 4.89 g of SrH2 and 5.28 g of H2O.
(a) How many moles of H2 can be produced from the given mass of SrH2?
This is a limiting reagent problem. I know that because an amount is given for both reactants.
SrH2 + 2H2O ==> Sr(OH)2 + 2H2
mols SrH2 = 4.89/molar mass SrH2.
mols H2O = 5.28/molar mass H2O.
Using the coefficients in the balanced equation, convert mols SrH2 to mols H2.
Do the same for mols H2O to mols H2.
It is likely the two values will not be the same which means one of them is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Now convert mols H2 to grams. g = mols x molar mass.
To calculate the number of moles of H2 that can be produced from the given mass of SrH2, we need to use the molar mass and stoichiometry of SrH2.
Step 1: Determine the molar mass of SrH2
SrH2 consists of one strontium (Sr) atom and two hydrogen (H) atoms. The atomic mass of Sr is 87.62 g/mol and the atomic mass of H is 1.01 g/mol. Adding these together, we get:
Molar mass of SrH2 = (1 × 87.62) + (2 × 1.01) = 89.63 g/mol
Step 2: Calculate the number of moles of SrH2
To calculate the number of moles of SrH2, we use the formula:
Number of moles = Mass / Molar mass
Given mass of SrH2 = 4.89 g
Number of moles of SrH2 = 4.89 g / 89.63 g/mol ≈ 0.0546 mol
Step 3: Apply stoichiometry
From the balanced equation, we know that 1 mole of SrH2 produces 2 moles of H2. Therefore, the number of moles of H2 that can be produced from the given mass of SrH2 is:
Number of moles of H2 = 0.0546 mol × 2 mol H2 / 1 mol SrH2 = 0.1092 mol H2
Therefore, 0.1092 moles of H2 can be produced from the given mass of SrH2.