If 3.21 g of BeC2O4 * 3H2O is heated to 220 C calculate the volume of the H2O(g) released, measured at 220 C and 735 mm Hg?
please help me
Convert g BeC2O4*3H2O to mols BeC2O4*3H2O.
Convert mols BeC2O4*3H2O to mols H2O.
Use PV = nRT to calculate volume of H2O.
Post your work if you get stuck.
3.21 g BeC2O4*3H2O x (1 mol BeC2O4*3H2O/238.14 g BeC2O4*3H2O) = 0.0134 mol BeC2O4*3H2O
0.0134 mol BeC2O4*3H2O x (3 mol H2O/1 mol BeC2O4*3H2O) = 0.0402 mol H2O
PV = nRT
V = (0.0402 mol H2O x 0.08206 L atm/mol K x 393.15 K)/(735 mm Hg)
V = 0.037 L H2O
To calculate the volume of H2O(g) released, measured at 220°C and 735 mm Hg, we need to follow a series of steps:
Step 1: Convert grams of BeC2O4 * 3H2O to moles of BeC2O4 * 3H2O.
First, calculate the molar mass of BeC2O4 * 3H2O:
Be: 9.012 g/mol
C: 12.011 g/mol
O: 16.00 g/mol
H: 1.008 g/mol
So, (9.012 * 1) + (12.011 * 2) + (16.00 * 4) + (1.008 * 3) = 151.064 g/mol
Now, divide the given mass (3.21 g) by the molar mass to get moles:
3.21 g / 151.064 g/mol = 0.02127 mol
Step 2: Convert moles of BeC2O4 * 3H2O to moles of H2O.
Since there are three water molecules per formula unit of BeC2O4 * 3H2O, we multiply the moles of BeC2O4 * 3H2O by three:
0.02127 mol * 3 = 0.0638 mol
Step 3: Use the ideal gas law, PV = nRT, to calculate the volume of H2O.
P = 735 mm Hg (convert to atm by dividing by 760 mm Hg/atm)
P = 735/760 = 0.967 atm
T = 220°C (convert to Kelvin by adding 273.15)
T = 220 + 273.15 = 493.15 K
R = 0.0821 L*atm/mol*K (universal gas constant)
Now, plug in the values into the equation:
V = (nRT) / P
V = (0.0638 mol * 0.0821 L*atm/mol*K * 493.15 K) / 0.967 atm
V ≈ 2.60 L
Therefore, the volume of H2O(g) released, measured at 220°C and 735 mm Hg, is approximately 2.60 liters.
To solve this problem, we need to follow the steps provided:
Step 1: Convert g BeC2O4*3H2O to mols BeC2O4*3H2O
To convert grams to moles, we need to use the molar mass of BeC2O4*3H2O.
The molar mass of BeC2O4*3H2O can be calculated as follows:
(1 Be x atomic mass of Be) + (2 C x atomic mass of C) + (4 O x atomic mass of O) + (3 H x atomic mass of H) + (6 O x atomic mass of O)
= (1 x 9.012) + (2 x 12.01) + (4 x 16.00) + (3 x 1.01) + (6 x 16.00)
= 9.012 + 24.02 + 64.00 + 3.03 + 96.00
= 196.062 g/mol
Now, we can calculate the number of moles of BeC2O4*3H2O:
Number of moles = (mass in grams) / (molar mass)
Number of moles = 3.21 g / 196.062 g/mol
Number of moles = 0.0164 mol
Step 2: Convert moles of BeC2O4*3H2O to moles of H2O
From the balanced equation of the reaction, we know that the ratio of moles of BeC2O4*3H2O to moles of H2O is 1:3.
Therefore, the number of moles of H2O released is:
Number of moles of H2O = (number of moles of BeC2O4*3H2O) x (3 moles of H2O / 1 mole of BeC2O4*3H2O)
Number of moles of H2O = 0.0164 mol x 3
Number of moles of H2O = 0.0492 mol
Step 3: Use PV = nRT to calculate the volume of H2O
The ideal gas law equation PV = nRT can be rearranged to solve for volume (V) in liters:
V = (nRT) / P
where:
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in kelvin)
P = pressure (in atm)
In this case, the pressure is given as 735 mmHg, which can be converted to atm:
Pressure (atm) = 735 mmHg / 760 mmHg/atm
Pressure (atm) = 0.966 atm
The temperature is given as 220 °C, which also needs to be converted to kelvin:
Temperature (K) = 273.15 + 220
Temperature (K) = 493.15 K
Now, we can substitute the values into the equation:
V = (0.0492 mol x 0.0821 L·atm/(mol·K) x 493.15 K) / 0.966 atm
V = 2.59 L
Therefore, the volume of H2O released, measured at 220 °C and 735 mmHg, is 2.59 liters.