Q22.77CP

Question

The key reaction (unbalanced) in the manufacture of synthetic cryolite for aluminum electrolysis is

$HF (g) + Al (OH)_{3} (s) + NaOH (a q) \to {Na}_{3} {AlF}_{6} (a q) + H_{2} O (ℓ)$

Assuming a $95.6%$ yield of dried, crystallized product, what mass (in ) of cryolite can be obtained from the reaction of $365 k g$ of ${Al(OH)}_{3}$ , $1.20 m^{3}$ of $50.0%$ by mass aqueous $N a O H (d = 1.53 g / m l)$ , and ${265m}^{3}$ of gaseous HF $305kP$ and $91 . 5^{°} C$ (Assume that the ideal gas law holds)

Step-by-Step Solution

Verified

Answer

The limiting reactant is $22951.7 mol$ .

The limiting reactant is HF $4443 mol$ .

The yield is $891 k g$ .

1Step 1: Concept introduction

The inorganic compound sodium aluminium hexafluoride (Na3AlF6) has the formula Na₃AlF₆. This white solid, discovered in 1799 by Peder Christian Abildgaard as the mineral cryolite, exists naturally and is widely employed in the industrial manufacturing of aluminium metal.

2Step 2: Analysing the intensity of the chemical reaction

Synthetic cryolite is made using the following balancing equation:

$6 H F (g) + A l (O H)_{3} (s) + 3 N a O H (a q) \to {Na}_{3} A l F_{6} (a q) + 6 H_{2} O (l)$

Let's write down everything we know:

$yield = 95.6 % m (A l {(O H)}_{3}) = 365 k g V (N a O H) = 1.20 m^{3} w (N a O H) = 50.0 % d (N a O H) = 1.53 g m L^{- 1} V (H F) = 265 m^{3} p (H F) = 305 k P a T (H F) = 91 . 5^{°} C$

By analysing the intensity of the chemical reaction for each component, we may compute the chemical quantities of each component and determine which one is the limiting reactant:

$ξ_{i} = \frac{n_{i}}{v_{i}}$

Therefore, we use $ξ_{i} = \frac{n_{i}}{v_{i}}$ to determine

3Step 3: Analysing the intensity of the chemical reaction

By analysing the intensity of the chemical reaction for each component, we may compute the chemical quantities of each component and determine which one is the limiting reactant:

$n (A l {(O H)}_{3}) = \frac{m (A l {(O H)}_{3})}{M (A l {(O H)}_{3})} (Al {(OH)}_{3}) = \frac{365 \cdot 10^{3} g}{78 gmol} n (Al {(OH)}_{3}) = 4679.5 mol (Al {(OH)}_{3}) = 1 ξ (Al {(OH)}_{3}) = 4679.5 n (N a O H) = \frac{m (N a O H)}{M (N a O H)} n (N a O H) = \frac{V (N a O H) \cdot d (N a O H) \cdot w (N a O H)}{M (N a O H)} n (N a O H) = 22951.7 mol$

Hence, the limiting reactant is $22951.7 mol$

4Step 4: Applying the ideal gas law

We must apply the ideal gas law to compute the amount of $H F$ :

$p V = n R T n = \frac{p V}{R T} n (H F) = \frac{305 \cdot 10^{3} Pa \cdot 265 m^{3}}{8.314 {JK}^{- 1} {mo}^{- 1} \cdot 364.65 K} n (H F) = 26660 mol v (H F) = 6 ξ (H F) = \frac{26660 mol}{6} = 4443 mol$

The limiting reactant is HF $4443 mol$ .

5Step 5: Computing Cryolite’s mass

We can now compute cryolite's mass.

$n (N a_{3} A l F_{6}) = \frac{1}{6} n (H F) n (N a_{3} A l F_{6}) = 4443 m o l M (N a_{3} A l F_{6}) = 209.94 g m o l m (N a_{3} A l F_{6}) = n (N a_{3} A l F_{6}) \cdot M (N a_{3} A l F_{6}) m (N a_{3} A l F_{6}) = 4443 m o l \cdot 209.94 g m o l^{- 1}$

$m (N a_{3} A l F_{6}) = 932763 g = 933 k g$

Of course, we must factor in the $95.6 \ %$ yield :

${(N a_{3} A l F_{6})}_{0.956} = m (N a_{3} A l F_{6}) \cdot \frac{95.6 %}{100 %} m {(N a_{3} A l F_{6})}_{0.956} = 933 k g \cdot 0.956 m {(N a_{3} A l F_{6})}_{0.956} = 891 k g$

Therefore, the yield is $891 k g$

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Q22.75CP

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