Respuesta :
From the periodic table:
molar mass of sodium = 23 gm
molar mass of Cr = 51.99 gm
molar mass of oxygen = 16 gm
molar mass of Na2Cr2O7 = 2(23) + 2(51.99) + 7(16) = 261.98 gm
number of moles in 7 gm = mass / molar mass = 7 / 261.98 = 0.0267 moles
1 mole of Na2Cr2O7 contains 7 moles of oxygen, therefore:
number of moles of oxygen in 0.0267 moles = 0.0267 x 7 = 0.1869 moles
To find the number of atoms, we multiply the number of moles by Avogadro's number as follow:
number of atoms = 0.1869 x 6.02 x 10^23 = 1.125138 x 10^23 atoms
molar mass of sodium = 23 gm
molar mass of Cr = 51.99 gm
molar mass of oxygen = 16 gm
molar mass of Na2Cr2O7 = 2(23) + 2(51.99) + 7(16) = 261.98 gm
number of moles in 7 gm = mass / molar mass = 7 / 261.98 = 0.0267 moles
1 mole of Na2Cr2O7 contains 7 moles of oxygen, therefore:
number of moles of oxygen in 0.0267 moles = 0.0267 x 7 = 0.1869 moles
To find the number of atoms, we multiply the number of moles by Avogadro's number as follow:
number of atoms = 0.1869 x 6.02 x 10^23 = 1.125138 x 10^23 atoms
[tex]\boxed{1.1255 \times {{10}^{23}}{\text{ atoms}}}[/tex] of oxygen are present in 7.00 g of sodium dichromate.
Further Explanation:
Given information:
Mass of sodium dichromate: 7.00 g
To calculate:
Number of oxygen atoms in 7.00 g of sodium dichromate
Steps to proceed:
I. First of all, moles of sodium dichromate that are present in 7.00 g of sodium dichromate are to be calculated. This is done with the help of equation (1) as mentioned below.
The formula to calculate moles of sodium dichromate [tex]\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}} \right)[/tex] is as follows:
[tex]{\text{Moles of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} = \dfrac{{{\text{Mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}}}{{{\text{Molar mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}}}[/tex] …… (1)
The mass of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}[/tex] is 7.00 g.
The molar mass of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}[/tex] is 261.97 g/mol.
Substitute these values in equation (1).
[tex]\begin{aligned}{\text{Moles of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} &= \frac{{{\text{7}}{\text{.00 g}}}}{{{\text{261}}{\text{.97 g/mol}}}} \\ &= 0.0267{\text{ mol}} \\ \end{aligned}[/tex]
II.Moles of oxygen that are present in 0.267 moles of sodium dichromate are to be calculated.
According to the chemical formula of sodium dichromate, it is evident that one mole of sodium dichromate contains two moles of sodium, two moles of chromium and seven moles of oxygen in it. Since one mole of sodium dichromate consists of seven moles of oxygen, moles of oxygen present in 0.267 moles of sodium dichromate can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of oxygen}} &= \left( {0.0267{\text{ mol N}}{{\text{a}}_2}{\text{C}}{{\text{r}}_4}{{\text{O}}_7}} \right)\left( {\frac{{7{\text{ mol O}}}}{{{\text{1 mol N}}{{\text{a}}_2}{\text{C}}{{\text{r}}_4}{{\text{O}}_7}}}} \right) \\&= 0.1869{\text{ mol O}} \\\end{aligned}[/tex]
III. Number of oxygen atoms that are present in 0.1869 moles of oxygen is to be calculated. This is done with the help of Avogadro’s law which states that one mole of a substance contains [tex]6.022 \times {10^{23}}{\text{ particles}}[/tex] . Such particles can be atoms, molecules, or formula units.
Since one mole of oxygen contains [tex]6.022 \times {10^{23}}{\text{ atoms}}[/tex], atoms of oxygen present in 0.1869 moles of oxygen can be evaluated as follows:
[tex]\begin{aligned}{\text{Number of oxygen atoms}} &= \left( {0.1869{\text{ mol}}} \right)\left( {\frac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{{\text{1 mol}}}}} \right) \\&= 1.1255 \times {10^{23}}{\text{ atoms}} \\\end{aligned}[/tex]
Hence, [tex]1.1255 \times {10^{23}}{\text{ atoms}}[/tex] of oxygen are present in 7.00 g of sodium dichromate.
Learn more:
- How many moles of Cl are present in 8 moles of [tex]{\text{CC}}{{\text{l}}_4}[/tex]? https://brainly.com/question/3064603
- Calculate the moles of ions in HCl solution: https://brainly.com/question/5950133
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: sodium dichromate, oxygen, mass, molar mass, 7.00 g, 1.1255*10^23 atoms, Avogadro’s law, one mole, atoms, molecules, particles.
