A 6.90-g sample of solid NiSO4·6H2O was heated such that the water turned to steam and was driven off. Assuming ideal behavior, what volume would that steam occupy at 1.00 atm and 100.0 °C?
To determine the volume of steam produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, we need to calculate the number of moles of steam produced from the given compound, NiSO4·6H2O.
To do this, we'll start by finding the number of moles of water (H2O) in the compound.
The molar mass of water (H2O) is:
2(1.01 g/mol for hydrogen) + 16.00 g/mol for oxygen = 18.02 g/mol
Since there are 6 water molecules (6H2O) in NiSO4·6H2O, we can calculate the number of moles of water:
6.90 g of NiSO4·6H2O × (1 mol H2O / 18.02 g H2O) = 0.383 mol H2O
Now we can calculate the number of moles of steam produced, assuming 1 mole of water vaporizes to form 1 mole of steam:
0.383 mol H2O × (1 mol steam / 1 mol H2O) = 0.383 mol steam
Next, we need to convert the temperature from Celsius to Kelvin:
100.0 °C + 273.15 = 373.15 K
Now that we have all the values, we can substitute them into the ideal gas law equation to solve for the volume (V):
V = (nRT) / P
V = (0.383 mol) × (0.0821 L·atm/mol·K) × (373.15 K) / 1.00 atm
V ≈ 12.26 L
Therefore, the volume of steam produced is approximately 12.26 liters at 1.00 atm and 100.0 °C.