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fredd [130]
2 years ago
10

If 12.2 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc

ed?
Chemistry
1 answer:
PilotLPTM [1.2K]2 years ago
6 0
<h2>Answer:  kilograms of cryolite ≈ 50.23 kg</h2>

Explanation:

<u><em>Write the balanced equation</em></u>

Let's start off by writing the reaction for the formation of cryolite:

Al₂O₃  +  6NaOH  +  12HF    →    2Na₃AlF₆  +  9H₂O

<em><u>Determine the limiting factor</u></em>

moles = mass ÷ molar mass

moles of Al₂O₃ = 12200 g ÷ 101.96 g/mol = 119.65 mol

moles of NaOH = 57400 g ÷  39.997 g/mol = 1435.11 mol

moles of HF = 57400 g ÷ 20.01 g/mol = 2868.57 mol

Since Al₂O₃ has the smallest number of moles in the reaction, it is the limiting factor. As such we will have to use the moles of Al₂O₃ to determine the moles of cryolite that will be produced.

<em><u>Find the mole ratio between the limiting factor and the cryolite</u></em>

The mole ratio of Al₂O₃ : Na₃AlF₆ based on the balanced equation is 1 : 2

∴ since the moles of Al₂O₃ = 119.65 mol

        the moles of Na₃AlF₆ = 119.65 mol × 2 = 239.3 mol

<u><em>Calculate the mass of the Cryolite</em></u>

Mass cryolite = moles × molar mass

                      = 239.3 mol × 209.9 g/mol

                      =  50229.07 g

      ∴ kilograms of cryolite ≈ 50.23 kg

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Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

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\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

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\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

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Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

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