Respuesta :
P₄O₁₀ + 6H₂O → 4H₃PO₄
The equation shows us that the molar ratio of
P₄O₁₀ : 6H₂O = 1:6
We also know that one mole of a substance contains 6.02 x 10²³ particles. We can use this to calculate the moles of water.
moles(H₂O) = (5.51 x 10²³) / (6.02 x 10²³)
= 0.92 mole
That means moles of P₄O₁₀ = 0.92 / 6
= 0.15
Each mole of P₄O₁₀ contains 4 moles of P.
moles(P) = 4 x 0.15 = 0.6 mol
Mr of P = 207 grams per mol
Mass of P = 207 x 0.6
= 124.2 grams
The equation shows us that the molar ratio of
P₄O₁₀ : 6H₂O = 1:6
We also know that one mole of a substance contains 6.02 x 10²³ particles. We can use this to calculate the moles of water.
moles(H₂O) = (5.51 x 10²³) / (6.02 x 10²³)
= 0.92 mole
That means moles of P₄O₁₀ = 0.92 / 6
= 0.15
Each mole of P₄O₁₀ contains 4 moles of P.
moles(P) = 4 x 0.15 = 0.6 mol
Mr of P = 207 grams per mol
Mass of P = 207 x 0.6
= 124.2 grams
[tex]$\boxed{{\text{18}}{\text{.582 g}}}$[/tex] of P is consumed if [tex]${\text{5}}{\text{.51}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}$[/tex] of [tex]${{\text{H}}_{\text{2}}}{\text{O}}$[/tex] is consumed in the reaction [tex]${{\text{P}}_4}{{\text{O}}_{10}}\left(s\right)+6{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)\to3{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}\left({aq}\right)$[/tex]
Further Explanation:
Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.
Consider the general reaction,
[tex]${\text{A}}+2{\text{B}}\to3{\text{C}}$[/tex]
Here,
A and B are reactants.
C is the product.
One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.
The given reaction is,
[tex]${{\text{P}}_4}{{\text{O}}_{10}}\left(s\right)+6{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)\to3{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}\left({aq}\right)$[/tex]
One mole of [tex]${{\text{P}}_4}{{\text{O}}_{10}}$[/tex] reacts with six moles of [tex]${{\text{H}}_{\text{2}}}{\text{O}}$[/tex] to produce three moles of [tex]${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$[/tex] . So the stoichiometric ratio between [tex]${{\text{P}}_4}{{\text{O}}_{10}}$[/tex] and [tex]${{\text{H}}_{\text{2}}}{\text{O}}$[/tex] is 1:6.
According to Avogadro law, one mole of any substance contains [tex]${\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}$[/tex] molecules.
So the number of moles of water is calculated as follows:
[tex]$\align{{\text{Number of moles of water}}{\bf&{=}}\left( {{\text{5}}{\text{.51}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}} \right)\left({\frac{{{\text{1 mol}}}}{{{\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}}\right)\cr&={\text{0}}{\text{.914978}}\;{\text{mol}}\cr&\approx{\bf{0}}{\bf{.92}}\;{\bf{mol}}\cr}$[/tex]
According to the stoichiometry of the reaction, 1 mole of [tex]${{\text{P}}_4}{{\text{O}}_{10}}$[/tex] reacts with 6 moles of [tex]${{\text{H}}_{\text{2}}}{\text{O}}$[/tex] to produce three moles of [tex]${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$[/tex] . So the number of moles of [tex]${{\text{P}}_4}{{\text{O}}_{10}}$[/tex] formed by 0.92 moles of [tex]${{\text{H}}_{\text{2}}}{\text{O}}$[/tex] is calculated as follows:
[tex]$\align{{\text{Moles of }}{{\text{P}}_4}{{\text{O}}_{10}}{\bf&{=}}\left({\frac{{{\text{0}}{\text{.92 mol}}}}{6}}\right)\cr&=0.1533\;{\text{mol}}\cr&\approx 0.15\;{\text{mol}}\cr}$[/tex]
1 mole of [tex]${{\text{P}}_4}{{\text{O}}_{10}}$[/tex] consists of 4 moles of P. So the moles of P can be calculated as follows:
[tex]$\align{{\text{Moles of P}}{\bf&{=}}\left({0.15\;{\text{mol}}}\right)\times4\cr&={\text{0}}{\text{.60 mol}}\cr}$[/tex]
The formula to calculate the mass of P is as follows:
[tex]${\text{Mass of P}}=\left({{\text{Moles of P}}}\right)\left({{\text{Molar mass of P}}}\right)$[/tex] …… (1)
The number of moles of P is 0.60 mol.
The molar mass of P is 30.97 g/mol.
Substitute these values in equation (1).
[tex]$\align{{\text{Mass of P}}&=\left({{\text{0}}{\text{.60 mol}}}\right)\left({\frac{{{\text{30}}{\text{.97 g}}}}{{{\text{1 mol}}}}}\right)\cr&={\bf{18}}{\bf{.582 g}}\cr}$[/tex]
Learn more:
1. How many moles of Cl are present in 8 moles of [tex]${\text{CC}}{{\text{l}}_4}$[/tex] ? https://brainly.com/question/3064603
2. Calculate the moles of ions in HCl solution: https://brainly.com/question/5950133
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: stoichiometry, P4O10, H2O, H3PO4, A, B, C, P, moles of P, Avogadro law, molecules, 0.60 mol, 0.15 mol, molar mass, number of moles, 18.582 g, 30.97 g/mol.
