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3 years ago

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A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter

containing water (the specific heat capacity of water is 4.18 J/g°C) initially at 21.1°C. The final temperature of water is 26.8°C. Calculate the mass of water in the calorimeter.
a. 1.7 g
b. 0.51 g
c. 150 g
d. 56 g
e. 61 g

1 answer:

Ira Lisetskai [31]3 years ago

5 0

Answer:

m H2O = 56 g

Explanation:

Q = mCΔT

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT)Al = (mCΔT)H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C

⇒ mH2O = 55.722 g ≅ 56 g

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Explanation:

<u>Step 1:</u> The balanced equation

2H2 + 02 → 2H20

<u>Step 2</u>: Given data

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<u>Step 3: </u>Calculate moles

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For 2 moles hydrogen consumed, we need 1 mole of oxygen.

This means oxygen is the limiting reagens and will be consumed completely. Hydrogen is the reactant in excess, there will remain 5.155 moles of hydrogen

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Now put all the given values in this expression, we get:

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