SOME ONE PLEASE HELP ME I NEED THIS SO BAD 30 POINTS AND WIL MARK BRAINLIST!!!
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The question is incomplete, here is the complete question:
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
<u>Answer:</u> The percent yield of water in the reaction is 46.85 %.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For NaOH:</u>
Given mass of NaOH = 77.0 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
- <u>For sulfuric acid:</u>
Given mass of sulfuric acid = 72.6 g
Molar mass of sulfuric acid = 98 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of NaOH and sulfuric acid follows:
By Stoichiometry of the reaction:
1 mole of sulfuric acid reacts with 2 moles of NaOH
So, 0.741 moles of sulfuric acid will react with = of NaOH
As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.
Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sulfuric acid produces 2 moles of water
So, 0.741 moles of sulfuric acid will produce = of water
Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 1.482 moles
Putting values in equation 1, we get:
To calculate the percentage yield of water, we use the equation:
Experimental yield of water = 12.5 g
Theoretical yield of water = 26.68 g
Putting values in above equation, we get:
Hence, the percent yield of water in the reaction is 46.85 %.
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.