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Elenna [48]
3 years ago
10

A 1.803 g sample of gypsum, a hydrated salt of calcium sulfate, CaSO4 is heated at a temperature greater than 170 degree Celsius

in a crucible until a constant mass is reached. The mass of anhydrous CaSO4 salt is 1.426g. Calculate the percent by mass of water in the hydrated calcium sulfate salt.
Chemistry
1 answer:
brilliants [131]3 years ago
6 0

Answer:

20.91% of the hydrated calcium sulfate salt is water

Explanation:

<u>Step1</u> : Calculate the mass of H2O that evaporated

A hydrated salt has a mass of 1.803g. After heating at 170° C, this means water will evaporate, the mass of the salt is 1.426g

The difference is water that evaporated.

⇒Mass H2O of hydration =1.803 g - 1.426g g = 0.377 g H2O

<u>Step 2</u>: Calculate the % of water in the hydrated calcium sulfate salt

⇒ mass of H2O = 0.377g

⇒ mass of calcium sulfate salt = 1.803g

⇒ % water in the calcium sulfate salt = (0.377 / 1.803) x 100% = 20.91%

20.91% of the hydrated calcium sulfate salt is water

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Answer:

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Explanation:

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<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>

It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>

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no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

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<u><em>Using cross multiplication:</em></u>

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

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where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

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To calculate the number of moles, we will be using the below formula

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