The reaction of 60.0 g of aluminum oxide with 30.0 g of carbon produced 22.5 g o

Question

The reaction of 60.0 g of aluminum oxide with 30.0 g of carbon produced 22.5 g of aluminum. What is the percent yield for this reaction? Al_0O_3 + 3C rightarrow + 3CO A) 44.9% B) 70.9% C) 50.1 % D) 31.8% E) 25.0% Calculate the volume occupied by 56.5 g, of argon gas at STP. A)31.7 L B) 1.380 L C) 22.4 L D) 34.6 L E) 1.270 L Determine the correct oxidation numbers for all three elements in Ca(ClO)_2 in the order that the elements are shown in the formula? A) -2, +3, -2 B) -2, +2, -1 C) +2, -2, +1 D) +2, +1, -2 E) +2, -3, +2 Given the specific heat for aluminum is 0.900 J/g degree C, how much heat is released when a 3.8 g sample of A1 cools from 450.0 degree C to 25 degree C. A) 1.7 kJ B) 1.5 kJ C) 60J D) 54 J E) 86J Calculate the volume occupied by 35.2 g of methane gas (CH_4) at 25 degree C and 1.0 atm. R = 0.08206 L atm/K mol. A) 4.5 L B) 11.2 L C) 49.2 L D) 53.7 L E) 0.0186 L

Explanation / Answer

(5)

Moles of alumina = mass / molar mass = 60 / 102 = 0.588 mol

Moles of Carbon = 30.0 / 12 = 2.50 mol

From the valanced equation,

1 mol of alumina needs 3 mol of C

then, 0.588 mol of alumina needs 3 * 0.588 = 1.76 mol of C ( < 2.50 mol)

Therefore, Alumina is limting reagent.

From the balanced equation,

1 mol alumina forms 2 mol of Al

then, 0.558 mol of alumina forms 0.588 *2 = 1.18 mol of Al

Now, theretical mass (yield) of Al = moles * molar mass = 1.18 * 27 = 31.8 g.

Actual yield = 22.5 g.

% yield = (actual yield / theretical yiled) * 100 = (22.5 / 31.8) * 100 = 70.8 %

(B)

(6)

At STP,

P = 1 atm

T = 273.15 K

R = 0.0821 L.atm.K-1.mol-1

n = mass / molar mass = 56.5 / 40 = 1.41 mol

V = ?

Ideal gas equation,

P V = n R T

V = 1.41 * 0.0821 * 273.15 / 1

V = 31.7 L

(A)

(7)

(D) + 2 , + 1 - 2

(8)

q = m * s (t2 - t1)

q = 3.8 * 0.900 * (25 - 450.0)

q = - 1453.5 J

q = - 1.45 kJ

(B)

(9)

V = n R T / P

V = m R T / M P

V = 35.2 * 0.08206 * 298.15 / (16 * 1.00)

V = 53.8 L

(D)

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