How many grams of CaH2 are needed to generate 54.5L of gas if the pressure of is 815 torr at 20C?
Use PV = nRT to calculate n, number of moles. moles and molar mass will get grams of H2. Then convert grams H2 to grams CaH2. Post your work if you get stuck.
The question is to find grams of CaH2. The balanced chemical equation is:
CaH2 + 2H2O --> Ca(OH)2 + 2H2
This mole ratio must be taken into account.
Find the moles of H2 as DrBob showed you, then multiply by 1/2 to get moles of CaH2. Finally convert moles of CaH2 to grams.
To find out how many grams of CaH2 are needed to generate 54.5L of gas at 815 torr and 20°C, we need to use the ideal gas law equation: PV = nRT.
Before we can use the equation, we need to convert some units:
- The pressure should be in atm, so we divide 815 torr by 760 torr/atm to get 1.071 atm.
- The volume should be in liters, which is already given as 54.5L.
- The temperature should be in Kelvin, so we add 273 to 20°C to get 293K.
Now we have the following information:
- P = 1.071 atm
- V = 54.5 L
- R = 0.0821 (constant for atm)
- T = 293 K
We can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the values:
n = (1.071 atm * 54.5 L) / (0.0821 L·atm/mol·K * 293 K)
n ≈ 2.114 mol
According to the balanced chemical equation for the reaction, 1 mol of CaH2 produces 1 mol of gas. Therefore, the number of moles of CaH2 needed is also 2.114 mol.
Next, we need to convert the number of moles of CaH2 to grams using its molar mass. The molar mass of CaH2 is:
Ca = 40.08 g/mol
H2 = 2.02 g/mol (2 hydrogen atoms)
Total molar mass of CaH2 = 40.08 g/mol + (2 * 2.02 g/mol) = 42.12 g/mol
Finally, we can calculate the mass of CaH2:
Mass (g) = n (mol) * molar mass (g/mol)
Mass (g) = 2.114 mol * 42.12 g/mol
Mass (g) ≈ 89.045 g
Therefore, approximately 89.045 grams of CaH2 are needed to generate 54.5L of gas at 815 torr and 20°C.