An 0.850 g sample of a mixture of LiH and CaH2 was reacted with H2O and the H2 gas produced, 1.20 L, was collected at STP.
What mass % of the original mixture was LiH?
To determine the mass % of LiH in the original mixture, we need to calculate the mass of LiH and CaH2 separately, and then find the mass % of LiH.
Step 1: Calculate the amount (in moles) of H2 produced
Since we know the volume of H2 gas collected and the conditions are given at STP (standard temperature and pressure), we can use the ideal gas law to calculate the number of moles of H2 gas produced.
PV = nRT
STP conditions:
Temperature (T) = 273.15 K
Pressure (P) = 1 atm
Convert the volume of H2 gas from liters to the unit used in the ideal gas law, which is moles:
1.20 L x (1 mol/22.4 L) = 0.0536 mol
Therefore, 0.0536 moles of H2 gas were produced.
Step 2: Determine the number of moles of LiH and CaH2 in the mixture
To find the number of moles of LiH and CaH2, we need to use the stoichiometry of the reaction between the metal hydrides and water.
The balanced equation for the reaction is:
2LiH + 2H2O -> 2LiOH + H2
or
CaH2 + 2H2O -> Ca(OH)2 + 2H2
Since we don't know the exact reactants, we need to consider the limiting reactant to determine the moles of LiH and CaH2.
Because 2 moles of H2 are produced for each mole of LiH or CaH2, and we know that 0.0536 moles of H2 were produced, we can divide this value by 2 to find the number of moles of metal hydrides.
0.0536 mol H2 / 2 = 0.0268 mol LiH or CaH2
Step 3: Calculate the mass of LiH and CaH2
To calculate the mass of LiH and CaH2, we need to use their respective molar masses.
The molar masses are:
LiH = 6.941 g/mol + 1.007 g/mol = 7.948 g/mol
CaH2 = 40.08 g/mol + 2.016 g/mol = 42.096 g/mol
Now we can calculate the mass of LiH and CaH2:
Mass of LiH = 0.0268 mol LiH x 7.948 g/mol = 0.213 g LiH
Mass of CaH2 = 0.0268 mol CaH2 x 42.096 g/mol = 1.129 g CaH2
Step 4: Calculate the mass % of LiH
To determine the mass % of LiH in the original mixture, divide the mass of LiH by the total mass of the mixture (LiH + CaH2) and multiply by 100.
Mass % of LiH = (0.213 g LiH / (0.213 g LiH + 1.129 g CaH2)) x 100
= (0.213 / (0.213 + 1.129)) x 100
= (0.213 / 1.342) x 100 = 15.9%
Therefore, the mass % of LiH in the original mixture is 15.9%.