\($\boxed{{\text{18}}{\text{.582 g}}}$\) of P is consumed if \(${\text{5}}{\text{.51}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}$\) of \(${{\text{H}}_{\text{2}}}{\text{O}}$\) is consumed in the reaction \(${{\text{P}}_4}{{\text{O}}_{10}}\left(s\right)+6{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)\to3{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}\left({aq}\right)$\)
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,
\(${\text{A}}+2{\text{B}}\to3{\text{C}}$\)
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,
\(${{\text{P}}_4}{{\text{O}}_{10}}\left(s\right)+6{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)\to3{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}\left({aq}\right)$\)
One mole of \(${{\text{P}}_4}{{\text{O}}_{10}}$\) reacts with six moles of \(${{\text{H}}_{\text{2}}}{\text{O}}$\) to produce three moles of \(${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$\) . So the stoichiometric ratio between \(${{\text{P}}_4}{{\text{O}}_{10}}$\) and \(${{\text{H}}_{\text{2}}}{\text{O}}$\) is 1:6.
According to Avogadro law, one mole of any substance contains \(${\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}$\) molecules.
So the number of moles of water is calculated as follows:
\($\align{{\text{Number of moles of water}}{\bf&{=}}\left( {{\text{5}}{\text{.51}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}} \right)\left({\frac{{{\text{1 mol}}}}{{{\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}}\right)\cr&={\text{0}}{\text{.914978}}\;{\text{mol}}\cr&\approx{\bf{0}}{\bf{.92}}\;{\bf{mol}}\cr}$\)
According to the stoichiometry of the reaction, 1 mole of \(${{\text{P}}_4}{{\text{O}}_{10}}$\) reacts with 6 moles of \(${{\text{H}}_{\text{2}}}{\text{O}}$\) to produce three moles of \(${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$\) . So the number of moles of \(${{\text{P}}_4}{{\text{O}}_{10}}$\) formed by 0.92 moles of \(${{\text{H}}_{\text{2}}}{\text{O}}$\) is calculated as follows:
\($\align{{\text{Moles of }}{{\text{P}}_4}{{\text{O}}_{10}}{\bf&{=}}\left({\frac{{{\text{0}}{\text{.92 mol}}}}{6}}\right)\cr&=0.1533\;{\text{mol}}\cr&\approx 0.15\;{\text{mol}}\cr}$\)
1 mole of \(${{\text{P}}_4}{{\text{O}}_{10}}$\) consists of 4 moles of P. So the moles of P can be calculated as follows:
\($\align{{\text{Moles of P}}{\bf&{=}}\left({0.15\;{\text{mol}}}\right)\times4\cr&={\text{0}}{\text{.60 mol}}\cr}$\)
The formula to calculate the mass of P is as follows:
\(${\text{Mass of P}}=\left({{\text{Moles of P}}}\right)\left({{\text{Molar mass of P}}}\right)$\) …… (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).
\($\align{{\text{Mass of P}}&=\left({{\text{0}}{\text{.60 mol}}}\right)\left({\frac{{{\text{30}}{\text{.97 g}}}}{{{\text{1 mol}}}}}\right)\cr&={\bf{18}}{\bf{.582 g}}\cr}$\)
Learn more:
1. How many moles of Cl are present in 8 moles of \(${\text{CC}}{{\text{l}}_4}$\) ? 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.
