metal hydrides react with water to form hydrogen gas and the metal hydroxide. How do you calculate the mass of hydrogen gas that can be prepared from 3.78 g of SrH2 and 3.5 g of H2O?
a) how many moles of H2 can be produced from given mass of SrH2?
b) How many grams of H2?
THANK YOU!!
This is a limiting reagent problem.
1. Write the equation and balance it.
SrH2 + 2H2O ==> Sr(OH)2 + 2H2 Check me out on that.
2a. Convert 3.78 g SrH2 to moles. moles = grams/molar mass.
2b. Convert 3.5 g H2O to moles.
3a. Using the balanced equation, convert moles SrH2 to moles H2.
3b. Using the balanced equation, convert moles H2O to moles H2.
3c. You have two answers for the same product and one of them MUST be wrong. The correct one, as in ALL limiting reagent problems, is the smaller one. The reactant producing that smaller number is the limiting reagent.
4. The moles from 3c x molar mass = grams produced.
To calculate the mass of hydrogen gas produced from 3.78 g of SrH2 and 3.5 g of H2O, we need to follow these steps:
a) Calculate the number of moles of H2 that can be produced from the given mass of SrH2.
1. Determine the molar mass of SrH2 (strontium hydride):
- The molar mass of Sr is 87.62 g/mol.
- The molar mass of H is 1.01 g/mol.
- Therefore, the molar mass of SrH2 is (87.62) + 2(1.01) = 89.64 g/mol.
2. Calculate the number of moles of SrH2:
- Divide the given mass of SrH2 by its molar mass:
Moles = Mass / Molar mass
Moles = 3.78 g / 89.64 g/mol = 0.0422 mol
3. Apply stoichiometry to determine the moles of H2 produced:
- From the balanced equation, we know that 1 mole of SrH2 produces 2 moles of H2.
- Therefore, the number of moles of H2 produced from 0.0422 mol of SrH2 is 2 * 0.0422 = 0.0844 mol.
b) Calculate the mass of H2 produced:
1. Determine the molar mass of H2:
- The molar mass of H2 is 2.02 g/mol (2 * 1.01).
2. Calculate the mass of H2:
- Multiply the number of moles of H2 by its molar mass:
Mass = Moles * Molar mass
Mass = 0.0844 mol * 2.02 g/mol = 0.170 g
Therefore, the mass of hydrogen gas that can be prepared from 3.78 g of SrH2 and 3.5 g of H2O is 0.170 g.