Chemistry: Xenon (xe) of mass 5.08 g reacts with fluorine to form 9.49 g of a xenon fluoride compound. what is the empirical formula of this compound?

Xenon (xe) of mass 5.08 g reacts with fluorine to

Ответ:

plb2007

29.09.2021

Chemistry

Approximately $0.121\; \rm mol$ .

Explanation:

Balance the equation

$\rm Cu\; (s)$ reacts with $\rm AgNO_3\; (aq)$ to produce $\rm Ag\; (s)$ and $\rm Cu(NO_3)_2\; (aq)$ .

$\rm ? \; Cu\, (s) + ?\; AgNO_3\, (aq) \to ?\; Ag\, (s) + ?\; Cu(NO_3)_2\, (aq)$ .

To balance this equation, start by setting the coefficient of the most complex species to . For example, $\rm Cu(NO_3)_2\, (aq)$ has more atoms in each of its formula unit than any other species in this reaction. Let the coefficient of

$\rm ? \; Cu\, (s) + ?\; AgNO_3\, (aq) \to ?\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)$ .

That formula unit of $\rm Cu(NO_3)_2\, (aq)$ would include:

$1 \times 1 = 1$ $\rm Cu$ atom, $1 \times 2= 2$ $\rm N$ atoms, and $1 \times 2 \times 3 = 6$ $\rm O$ atoms.

The other product, $\rm Ag\; (s)$ , contains neither $\rm Cu$ atoms nor $\rm N$ atoms. Therefore, the product side would include exactly

one $\rm Cu$ atom, two $\rm N$ atoms, andsix $\rm O$ atoms.

Atoms are conserved in a chemical reaction. The reactant side shall also include:

$1 \times 1 = 1$ $\rm Cu$ atom, $1 \times 2= 2$ $\rm N$ atoms, and $1 \times 2 \times 3 = 6$ $\rm O$ atoms.

Among the reactants, $\rm AgNO_3\, (aq)$ is the only source of $\rm N$ atoms. Each formula unit of

Similarly, the coefficient of $\rm Cu$ would be 1/ 1 = 1 .

$\rm 1 \; Cu\, (s) + 1\; AgNO_3\, (aq) \to ?\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)$ .

There would be exactly $\rm Ag$ atom on the reactant side (from $\rm AgNO_3\, (aq)$ .) The products should also include

Hence the equation:

$\rm 1 \; Cu\, (s) + 1\; AgNO_3\, (aq) \to 2\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)$ .

Calculate n(Cu(NO₃)₂ (aq))

The ratio between the coefficients of $\rm Cu(NO_3)_2\; (aq)$ and $\rm Ag\, (s)$ is:

$\displaystyle \frac{n(\mathrm{Cu(NO_3)_2\, (aq)})}{n(\mathrm{Ag\, (s)})} = \frac{1}{2}$ .

Rearrange this equation to obtain:

$\displaystyle n(\mathrm{Cu(NO_3)_2\, (aq)}) = \frac{1}{2} \, n(\mathrm{Ag\, (s)})$ .

In other words, $n(\rm Cu(NO_3)_2\; (aq))$ (the number of moles of $\rm Cu(NO_3)_2\; (aq)$ that is produced) can be found from $n(\mathrm{Ag\, (s)})$ (the number of moles of $\rm Ag\, (s)$ that is produced.)

On the other hand, $n(\rm Ag\; (aq))$ can be found from $m(\rm Ag\; (s))$ (the mass of $\rm Ag\, (s)$ that is produced) using $M(\rm Ag)$ (the formula mass of $\rm Ag$ .)

Look up relative atomic mass data of $\mathrm{Ag}$ on a modern periodic table:

$\rm Ag$ : 107.868

Therefore, the relative atomic mass of $\rm Ag\, (s)$ would be:

$M(\rm Ag) = 107.868\; \rm g \cdot mol^{-1}$ .

Calculate $n(\rm Ag\; (s))$ :

$\begin{aligned}& n({\rm Ag\, (s)}) \\ &= \frac{m(\mathrm{Ag\, (s)})}{M(\mathrm{Ag})}\\ &=\frac{2.60\; \rm g}{107.868 \; \rm g \cdot mol^{-1}} \\ &\approx 0.0241035\; \rm mol \end{aligned}$ .

Calculate $n(\rm Cu(NO_3)_2\; (aq))$ :

$\displaystyle n(\mathrm{Cu(NO_3)_2\, (aq)}) = \frac{1}{2} \, n(\mathrm{Ag\, (s)}) \approx 0.0121\; \rm mol$ .

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Xenon (xe) of mass 5.08 g reacts with fluorine to form 9.49 g of a xenon fluoride compound. what is the empirical formula of this compound?

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