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 percentage of LiH in the original mixture, we can use stoichiometry and the ideal gas law. Here are the steps to calculate it:
Step 1: Calculate the number of moles of H2 gas produced.
According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
At STP (Standard Temperature and Pressure), the pressure is 1 atm, and the temperature is 273 K. The volume of gas collected is 1.20 L.
Using the ideal gas law equation, we can rearrange it to solve for n (number of moles) as follows:
n = PV / RT
Plug in the given values:
n = (1 atm) * (1.20 L) / [(0.0821 L * atm / K * mol) * (273 K)]
n ≈ 0.055 mol
Step 2: Determine the molar ratio between LiH and H2.
The balanced chemical equation for the reaction between the LiH or CaH2 with H2O is:
LiH + H2O -> LiOH + H2
From the equation, we can see that 1 mole of LiH reacts to produce 1 mole of H2.
Step 3: Calculate the number of moles of LiH.
Since we know the molar ratio is 1:1 between LiH and H2, the number of moles of LiH is also 0.055 mol.
Step 4: Calculate the molar mass of LiH.
The molar mass of LiH can be calculated by adding the atomic masses of lithium (Li) and hydrogen (H).
Molar mass of LiH = (atomic mass of Li) + (atomic mass of H)
Molar mass of LiH = (6.94 g/mol) + (1.01 g/mol)
Molar mass of LiH ≈ 7.95 g/mol
Step 5: Calculate the mass of LiH in the original mixture.
Mass of LiH = number of moles of LiH * molar mass of LiH
Mass of LiH ≈ 0.055 mol * 7.95 g/mol
Mass of LiH ≈ 0.438 g
Step 6: Calculate the mass percentage of LiH in the original mixture.
Mass % = (mass of LiH / mass of original mixture) x 100
Mass % = (0.438g / 0.850g) x 100
Mass % ≈ 51.53%
Therefore, the mass percentage of LiH in the original mixture is approximately 51.53%.
To find the mass % of LiH in the original mixture, we need to determine the mass of LiH and CaH2 separately and then calculate the percentage.
First, let's start by finding the number of moles of H2 produced using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Since the reaction took place at STP (Standard Temperature and Pressure), we can use the values:
P = 1 atm
V = 1.20 L
T = 273.15 K
R = 0.0821 L.atm/mol.K (gas constant)
Using the ideal gas law equation, we can rearrange it to solve for n (number of moles):
n = (PV) / (RT)
Substituting the values:
n = (1 atm * 1.20 L) / (0.0821 L.atm/mol.K * 273.15 K)
Now we have the number of moles of H2 gas produced.
Next, we need to calculate the molar mass of LiH and CaH2 to determine the ratio of moles of LiH to the total moles of hydrides in the mixture.
The molar mass of LiH is 7.95 g/mol (6.94 g/mol for Li + 1.01 g/mol for H) and the molar mass of CaH2 is 42.09 g/mol (40.08 g/mol for Ca + 2.01 g/mol for H).
Now we can calculate the moles of LiH and CaH2 in the mixture:
moles of LiH = (mass of LiH) / (molar mass of LiH)
moles of CaH2 = (mass of CaH2) / (molar mass of CaH2)
To find the mass of each compound, we need to use the given mass of the mixture (0.850 g) and the percentage of LiH or CaH2 in the mixture.
Let's assume x is the mass % of LiH in the mixture. This means the mass % of CaH2 will be (100 - x).
Therefore,
mass of LiH = (x / 100) * 0.850 g
mass of CaH2 = [(100 - x) / 100] * 0.850 g
Now we can calculate the moles of each compound:
moles of LiH = (x / 100) * 0.850 g / molar mass of LiH
moles of CaH2 = [(100 - x) / 100] * 0.850 g / molar mass of CaH2
Now, we need to find the ratio of moles of LiH to the total moles of hydrides in the mixture:
ratio =[ moles of LiH / (moles of LiH + moles of CaH2) ]
Finally, we can calculate the mass % of LiH in the initial mixture:
mass % of LiH = ratio * 100
By plugging in the given values and following these steps, we can determine the mass % of LiH in the original mixture.
work it in two parts. let V be the volume of H2 produced in the CaH2 reaction, and 1.2-V be the volume in the LiH reaction. Let m be mass in the CaH2 reaction, and .850g -m be the mass in the LiH reaction.
You will get two equations, one unknown out of this.
CaH2 reaction:
CaH2 + 2H2O = Ca(OH)2 + 2H2
H2: P:1 atm T: 273K V: V
PV=nRT
nCa=PV/RT
n= (1*V)/(0.08206*273)
n= you do it.
LiH reaction:
same method, solve for NLi
you will get N in terms of V
now convert each of the N's to grams (by multiplying each by formaula masses of the hydrides), and set each to a funcion of m.
you should have time then to find the ratios of the two masses, and solve for percent. A bit of algebra is required.
LiH + H2O ==> LiOH + H2
CaH2 + 2H2O ==> Ca(OH)2 + 2H2
You must write two equations and solve them simultaneously. NOTE; ALL OF THESE NUMBERS ARE ESTIMATES; YOU SHOULD RECALCULATE ALL OF THEM.
Let Y stand for grams LiH.
and Z stand for grams CaH2.
Then equation 1 is Y + Z + 0.850
How many mols is 1.50 L @ STP. That's 1.50/22.4 = aprox 0.067
For the second equation you want to write an equation that equates mols H2 from LiH and mols H2 from CaH2 = 0.067
Below I will use mm for molar mass.
So mols H2 from LiH will be
[(Y/mm LiH)] x [(1 mol H2/1 mol LiH)]
and mol H2 from CaH2 will be
[(Z/mm CaH2)] x [2 mols H2/1 mol CaH2)]
Solve for Y = grams LiH.
Then % LiH = (g LiH/mas sample)*100
Post your work if you get stuck.