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Naily [24]
3 years ago
6

Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?

Chemistry
1 answer:
s2008m [1.1K]3 years ago
4 0

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

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

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

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Sodium phosphate is added to a solution that contains 0.0070 M aluminum nitrate and 0.052 M calcium chloride. The concentration
boyakko [2]

Answer:

The answer to the question is;

The first ion to precipitate out is the Al³⁺ ion and the concentration of the Al³⁺ ion when the Ca²⁺ ion begins to precipitate is 1.12 × 10⁻⁵ M.

Explanation:

To solve the question, we note that

aluminum nitrate, Al(NO₃)₃ will dissociate as follows

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Therefore when sodium phosphate is added to a solution that contains aluminum nitrate  we have the following system  of aluminium phosphate which is

AlPO₄(s) ⇄ Al³⁺(aq) + PO₄³⁻(aq)

The solubility product for the above reaction is

Ksp = [Al³⁺][PO₄³⁻] = 9.84×10⁻²¹

The solubility product for calcium phosphate is expressed as

Ca₃(PO₄)₂(s) ⇄ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)

With Ksp =  [Ca²⁺]³[PO₄³⁻]² = 2.07×10⁻³³

From the solubility product, we can find the concentration of [PO₄³⁻] at which precipitation starts as follows

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[PO₄³⁻] = \frac{K_{sp}}{[Al^{3+}]}= 9.84×10⁻²¹ / 0.007 = 1.406×10⁻¹⁸ M

The phosphate concretion for Ca²⁺ when precipitation starts is

[PO₄³⁻] =\sqrt{\frac{K_{sp}}{[Ca^{2+}]^2}}  = \sqrt{\frac{2.07\times10^{-33}}{[0.052]^2}} = 8.75×10⁻¹⁶ M

(Aluminium phosphate precipitates out first)

The reaction favors the precipitation of the aluminum phosphate first due to the lower concentration of the [PO₄³⁻]  ions in the [Al³⁺][PO₄³⁻] system which  is lower than the relative [PO₄³⁻] in the [Ca²⁺]³[PO₄³⁻]².

Therefore, the more sodium phosphate added serves to precipitate the remaining aluminium phosphate.

The process continues and the concentration of Al³⁺ decreases as more precipitates form. The process continues until the equilibrium conditions satisfies the precipitation threshold level for the calcium phosphate system concentration whereby the concentration of the Al³⁺ in the solution is given by.

[Al³⁺] = \frac{K_{sp}}{[PO_4^{3-}]} = \frac{9.84\times 10^{-21}}{8.75\times 10^{-16}}  = 1.12 × 10⁻⁵ M

Therefore the concentration of this aluminium ion when the calcium ion begins to precipitate =  1.12 × 10⁻⁵ M.

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