Respuesta :
1. Calculate the molar masses of each compound:
[tex]C_2H_6 = 12\cdot 2 + 1\cdot 6 = 30\ g/mol[/tex]
[tex]CCl_3F = 12\cdot 1 + 35.5\cdot 3 + 19\cdot 1 = 137.5\ g/mol[/tex]
2. Both masses are referred to one mole of the substances. We can do a ratio between them and the 3 g of [tex]CCl_3F[/tex]:
[tex]\frac{30\ g\ C_2H_6}{137.5\ g\ CCl_3F} = \frac{x}{3\ g\ CCl_3F}\ \to\ x = \bf 0.654\ g\ C_2H_6[/tex]
The mass of [tex]${\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}$[/tex] is 0.654 g
Mole is a measure of the amount of substance. It is defined as the mass of a substance that has the same number of fundamental units as there are atoms in 12 g of carbon-12. Such fundamental units can be atoms, molecules or formula units.
Step1: Initially we have to calculate the moles of trichlorofluoromethane [tex]$\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$[/tex]
The formula to calculate the moles of trichlorofluoromethane [tex]$\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$[/tex] is as follows:
[tex]${\text{Moles of CC}}{{\text{l}}_{\text{3}}}{\text{F}}=\frac{{{\text{Given mass of CC}}{{\text{l}}_{\text{3}}}{\text{F}}}}{{{\text{Molar mass of CC}}{{\text{l}}_{\text{3}}}{\text{F}}}}$[/tex] …… (1)
The given mass of [tex]$\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$[/tex] is 3 g.
The molar mass of [tex]$\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$[/tex] is 137.34 g/mol.
Substitute these values in equation (1)
[tex]$\align{{\text{Moles of CC}}{{\text{l}}_{\text{3}}}{\text{F}}{\bf&{=}}\left({3\;{\text{g}}}\right)\left({\frac{{{\text{1}}\;{\text{mol}}}}{{{\text{137}}{\text{.34}}\;{\text{g}}}}}\right)\cr&=0.02184\;{\text{mol}}\cr}$[/tex]
The number of moles of trichlorofluoromethane [tex]$\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$[/tex] is equal to the number of moles of ethane [tex]$\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$[/tex] and that is 0.02184 moles.
Step2: Calculate the mass of ethane [tex]$\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$[/tex] from the moles of ethane [tex]$\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$[/tex] .
The formula to calculate the moles of ethane [tex]$\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$[/tex] is as follows:
[tex]${\text{Moles of C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}=\frac{{{\text{Given mass of C}}{& _{\text{2}}}{{\text{H}}_{\text{6}}}}}{{{\text{Molar mass of C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}}$[/tex] …… (2)
The moles of [tex]$\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$[/tex] is 0.02184 mol.
The molar mass of [tex]$\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$[/tex] is 30 g/mol.
Substitute these values in equation (2)
[tex]$0.02184\;{\text{mol}}{\bf{=}}\left({{\text{Given mass of C}}{ & _{\text{2}}}{{\text{H}}_{\text{6}}}}\right)\left({\frac{{{\text{1}}\;{\text{mol}}}}{{{\text{137}}{\text{.34}}\;{\text{g}}}}}\right)$[/tex]
(3)
Rearrange the equation (3) to calculate the mass of [tex]$\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$[/tex]
[tex]$\align{{\text{Given mass of }}{{\text{C}}_2}{{\text{H}}_6}&=\frac{{0.02184\;{\text{mol}}}}{{{\text{137}}{\text{.34}}\;{\text{g/mol}}}}\cr&=0.654\;{\text{g}}\cr}$[/tex]
Hence, the mass of [tex]$\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$[/tex] is 0.654 g.
Learn more:
1. Bond energy of H-H bond in the given reaction: https://brainly.com/question/7213980
2. The main purpose of conducting experiments: https://brainly.com/question/5096428
Answer details:
Grade: High School
Subject: Chemistry
Chapter: Mole concept
Keywords: stoichiometry, C2H6, CCL3F, moles, mass, molar mass, 0.02184 mol., 30 g/mol, 3 g, 137.34 g/mol and 0.654 g.
