Respuesta :
Answer:
The maximum mass of phosphoric acid that can be formed is 34.2 g
The molecular formula of the limiting reagent is P₄O₁₀
The mass of the excess reagent (water) that remains after the reaction is complete is 3.76 g
Explanation:
Hi there!
Let´s write the balanced chemical equation:
P₄O₁₀ + 6 H₂O → 4 H₃PO₄
The molar masses of the reactants and the product are the following:
P₄O₁₀ = 284 g
H₂O = 16 g
H₃PO₄ = 98 g
The number of moles of the reactants we have are the following:
24.8 g P₄O₁₀ · 1 mol/ 284 g = 0.0873 mol P₄O₁₀
13.2 g H₂O · 1 mol/ 18 g = 0.733 mol H₂O
We know from the chemical equation that 1 mol of P₄O₁₀ reacts with 6 mol of H₂O, then, 0.0873 mol P₄O₁₀ will react with:
0.0873 mol P₄O₁₀ · 6 mol H₂O/ 1 mol P₄O₁₀ = 0.524 mol H₂O
Then, water is in excess because we have 0.733 mol and only 0.524 mol will react. Thus, the limiting reagent is P₄O₁₀
The number of moles of water that will remain after the reaction is complete is the following:
0.733 mol - 0.524 mol = 0.209 mol
Let´s convert that amount of water into mass units
0.209 mol · 18 g / mol = 3.76 g
3.76 g of water remains after the reaction is complete.
Assuming that the reaction has a yield of 100% since 1 mol of P₄O₁₀ produces 4 mol of phosphoric acid, 0.0873 mol P₄O₁₀ will produce
( 0.0873 mol P₄O₁₀ · 4 mol H₃PO₄/ mol P₄O₁₀ ) 0.349 mol phosphoric acid.
Then, the maximum mass of phosphoric acid that can be formed is the following:
0.349 mol H₃PO₄ · 98 g / 1 mol = 34.2 g
The maximum mass of phosphoric acid that can be formed is 34.2 g.
The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex]
The excess reagent that remains after the reaction is complete is 3.76 g.
What is phosphoric acid?
Phosphoric acid (H3PO4), also known as orthophosphoric acid, is the most important oxygen acid of phosphorus.
It is a crystalline solid and is used to generate fertilizer phosphate salts.
Given,
The balanced chemical equation is
[tex]P_4O_1_0 + 6 H_2O =4 H_3PO_4[/tex]
The molar mass of the following
[tex]\bold{P_4O_1_0}[/tex] = 284 g
[tex]\bold{H_2O}[/tex] = 16 g
[tex]\bold{H_3PO_4}[/tex] = 98 g
Calculating the number of moles
[tex]\dfrac{\bold{P_4O_1_0}}{284 g} = 0.0873\; mol/P_4O_1_0[/tex]
[tex]\dfrac{\bold{13.2\;g\;H_2O}}{18 g} = 0.733\; mol/P_4O_1_0[/tex]
From equation, 1 mol of [tex]\bold{P_4O_1_0}[/tex] reacts with 6 mol of water, then for 0.0873 mol of [tex]\bold{P_4O_1_0}[/tex] will react
[tex]0.0873\; mol\; P_4O_1_0 \times 6 mol H_2O/ 1 molP_4O_1_0 = 0.524\; mol\; H_2O[/tex]
The number of moles of water remaining are:
0.733 mol - 0.524 mol = 0.209 mol
Convert the amount of water into mass unit
0.209 mol · 18 g / mol = 3.76 g
The number of moles of water remaining is 3.76 g
Assuming that the reaction has a yield of 100%
Since 1 mol of [tex]\bold{P_4O_1_0}[/tex] produces 4 mol of phosphoric acid, 0.0873 mol [tex]\bold{P_4O_1_0}[/tex] will produce:
[tex]0.0873\; mol\; P_4O_1_0 \times 4 mol H_3PO_4/ molP_4O_1_0 = 0.349\; mol\; of\;phosphoric\;acid[/tex]
Then, the maximum mass of phosphoric acid that can be formed is
[tex]0.349\; mol\; H_3PO_4 \times 98\;g/ mol = 34.2\; g[/tex]
Thus, The maximum mass of phosphoric acid is 34.2 g, The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex], and the excess reagent that remains after the reaction is complete is 3.76 g.
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