Round to 4 significant figures. 0.007062
1 answer:
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Answer:
a) 157.5 grams of aluminum.
b) 1 mol
c) 9 g
Explanation:
The reaction is :
As per balanced equation
a) 3 moles of hydrogen will be produced from two moles of aluminium.
The atomic mass of aluminium = 27
therefore
3X2 grams of hydrogen is produced from 2 X 27 grams of Al
1 gram of hydrogen will be produced from g
therefore 17.5 will be produced from = 9X 17.5 = 157.5 grams of aluminum.
b) as per balanced equation three moles or six gram of hydrogen is produced from 6 moles of NaOH.
Therefore 1 g of hydrogen will be produced from =
or 1 gram will be prepared from = 1 mole
c) from balanced equation three moles are produced from two moles of Al (27X2 = 54 g).
thus from 54 grams gives 6 grams of hydrogen
1 grams will give =
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
Al(NO₃)₃ → Al³⁺ (aq) + 3NO₃⁻ (aq)
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
The phosphate concretion for Al³⁺ when precipitation starts is
[PO₄³⁻] = = 9.84×10⁻²¹ / 0.007 = 1.406×10⁻¹⁸ M
The phosphate concretion for Ca²⁺ when precipitation starts is
[PO₄³⁻] = =
= 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⁺] = = 1.12 × 10⁻⁵ M
Therefore the concentration of this aluminium ion when the calcium ion begins to precipitate = 1.12 × 10⁻⁵ M.