Respuesta :
The solution for the problem is:746 mmHg of F2 would respond wholly with 746 x (1/3) = 249 mmHg of Cl2, nonetheless there is more Cl2 current than that, so Cl2 is in excess and F2 is the limiting reactant.
n = PV / RT = (746 mmHg F2) x (2.15 L) / ((62.36367 L mmHg/K mol) x (298 K)) = 0.0863 mol F2
(0.0863 mol F2) x (2 mol ClF3 / 3 mol F2) x (92.4482 g ClF3/mol) = 5.3210 g ClF3
n = PV / RT = (746 mmHg F2) x (2.15 L) / ((62.36367 L mmHg/K mol) x (298 K)) = 0.0863 mol F2
(0.0863 mol F2) x (2 mol ClF3 / 3 mol F2) x (92.4482 g ClF3/mol) = 5.3210 g ClF3
The limiting reactant in given reaction is [tex]\boxed{{{\text{F}}_{\text{2}}}}[/tex] and theoretical yield of [tex]{\text{Cl}}{{\text{F}}_{\text{3}}}[/tex] is [tex]\boxed{5.32{\text{ g}}}[/tex].
Further explanation:
Limiting reagent is completely consumed in any chemical reaction and its amountdecides the amount of product formed after chemical reaction.
Stoichiometry is used to determine amount of species by using relationships between reactants and products.
Given reaction is as follows:
[tex]{\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right) + 3{{\text{F}}_{\text{2}}}\left( {\text{g}} \right) \to 2{\text{Cl}}{{\text{F}}_{\text{3}}}[/tex]
According to the stoichiometry of the above reaction, one mole of [tex]{\text{C}}{{\text{l}}_2}[/tex] reacts with three moles of [tex]{{\text{F}}_2}[/tex] and two moles of [tex]{\text{Cl}}{{\text{F}}_{\text{3}}}[/tex]. Since pressure and number of moles are directly related to each other, same stoichiometry can be applied to their pressure terms. So partial pressure of [tex]{\text{C}}{{\text{l}}_2}[/tex] required to react completely with 746 mm Hg of [tex]{{\text{F}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned} {\text{Partial pressure of C}}{{\text{l}}_{\text{2}}} &= \frac{1}{3}\left( {746{\text{ mm Hg}}} \right) \\ &= 248.6{\text{ mm Hg}} \\ &\approx {\text{249 mm Hg}} \\ \end{aligned}[/tex]
But partial pressure of [tex]{\text{C}}{{\text{l}}_2}[/tex] is 337 mm Hg. This indicates its moles are present in excess amount and therefore [tex]{{\text{F}}_{\text{2}}}[/tex] acts as limiting reactant.
Ideal gas law is the equation of state for any hypothetical gas. Itsmathematical form is as follows:
[tex]{\text{PV}} = {\text{nRT}}[/tex] …… (1)
Where,
P is the pressure of ideal gas.
V is the volume of ideal gas.
T is the absolute temperature of ideal gas.
n is the number of moles of gas.
R is the universal gas constant.
Rearrange equation (1) for n.
[tex]{\text{n}} = \dfrac{{{\text{PV}}}}{{{\text{RT}}}}[/tex] …… (2)
Substitute 746 mm Hg for n, 2.15 L for V, 298 K for T and 62.36367 L.mm Hg/K.mol for R in equation (2).
[tex]\begin{aligned} {\text{n}} &= \frac{{\left( {746{\text{ mm Hg}}} \right)\left( {2.15{\text{ L}}} \right)}}{{\left( {62.36367{\text{ L}}{\text{.mm Hg/K}}{\text{.mol}}} \right)\left( {298{\text{ K}}} \right)}} \\ &= 0.0863{\text{ mol}} \\ \end{aligned}[/tex]
Stoichiometry of given reaction indicates that three moles of [tex]{{\text{F}}_{\text{2}}}[/tex] can form two moles of [tex]{\text{Cl}}{{\text{F}}_3}[/tex]. Therefore number of moles of [tex]{\text{Cl}}{{\text{F}}_3}[/tex] formed by 0.0863 moles of [tex]{{\text{F}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned} {\text{Moles of Cl}}{{\text{F}}_{\text{3}}} &= \left( {0.0863{\text{ mol }}{{\text{F}}_2}} \right)\left( {\frac{{2{\text{ mol Cl}}{{\text{F}}_3}}}{{3{\text{ mol }}{{\text{F}}_2}}}} \right) \\ &= 0.0575{\text{ mol}} \\ \end{aligned}[/tex]
The formula to calculate mass of [tex]{\text{Cl}}{{\text{F}}_3}[/tex] is as follows:
[tex]{\text{Mass of Cl}}{{\text{F}}_{\text{3}}} = \left( {{\text{Moles of Cl}}{{\text{F}}_3}} \right)\left( {{\text{Molar mass of Cl}}{{\text{F}}_3}} \right)[/tex] …… (3)
Substitute 0.0575 mol for moles of [tex]{\text{Cl}}{{\text{F}}_3}[/tex] and 92.448 g/mol for molar mass of in equation (3).
[tex]\begin{aligned} {\text{Mass of Cl}}{{\text{F}}_{\text{3}}} &= \left( {0.0575{\text{ mol}}} \right)\left( {92.448{\text{ g/mol}}} \right) \\ &= 5.32{\text{ g}} \\ \end{aligned}[/tex]
Hence theoretical yield of [tex]{\text{Cl}}{{\text{F}}_3}[/tex] is 5.32 g.
Learn more:
- Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603
- How many grams of potassium was in the fertilizer? https://brainly.com/question/5105904
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: limiting reagent, ClF3, Cl2, F2, stoichiometry, mass of ClF3, moles of ClF3, molar mass of ClF3, 0.0863 moles, 5.32 g.
