Question

A 6.30 g 6.30 g sample of solid CaSO 4 ⋅ 2 H 2 O CaSO4⋅2H2O 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 1.00 atm and 100.0 °C?

Answer 1

To find the volume of steam produced when hydrate calcium sulfate is heated, we first calculate the moles of water vaporized and then use the Ideal Gas Law, considering the conditions of 1.00 atm pressure and 100.0 °C temperature.

Step-by-step explanation:

The question involves the calculation of the volume of steam generated when a sample of hydrated calcium sulfate (CaSO4·2H2O) is heated to drive off the water content. The steam will occupy a certain volume under the given conditions of temperature and pressure, which can be computed by using the Ideal Gas Law.

First, we need to know the amount (in moles) of water that is released upon heating. Since the formula of the hydrate indicates that there are two moles of water for each mole of CaSO4, we can calculate the moles of water based on the mass of the CaSO4·2H2O sample. With the molar mass of CaSO4·2H2O and the molar mass of water (H2O), this calculation gives us the number of moles of water that were present as steam.

To find the volume of this steam, we apply the Ideal Gas Law, PV = nRT, where P is the pressure (1.00 atm), V is the volume we want to find, n is the number of moles of steam, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature (100.0 °C or 373.15 K).

After plugging in the known values and solving for V, we have the volume of steam at the specified conditions of 1.00 atm and 100.0 °C.

Answer 2

The volume of steam V = 0.0107
`m^(3)`

Step-by-step explanation:

Mass m = 6.3 gm = 0.0063 kg

Pressure = 1 atm = 101.325 k pa

Temperature = 100 °c = 373 K

From ideal gas equation,

P V = m R T --------- (1)

Put all the values in the above equation we get

101.325 × V = 0.0063 × 0.462 × 373

⇒ V = 0.0107
`m^(3)`

This is the volume of steam.