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3 years ago

What is the percent composition by mass of aluminum in Al2(SO4)3 (gram-formula mass = 342 grams/mole)?

2 answers:

8 0

The atomic mass of Aluminum is around 27 amu. There are two aluminum atoms, so it takes 54 amu. The gram-formula states that 1 mole of this compound has 342 grams. By definition, this means that the total amu of this compound is 342. Therefore, we divide 54 by 342 to get around 15.8 %.

Fantom [35]3 years ago

3 0

Answer:

The correct answer is option (2).

Explanation:

Molar mass of aluminum sulfate = 342 g/mol

Atomic mass of aluminium = 26.98 g/mol

In a molecule of aluminum sulfate there are 2 atoms of aluminium.

Mass of aluminum in aluminium sulfate:

$2\times 26.98 g/mol=53.96 g/mol$

Percent composition by mass of aluminum in $Al_2(SO_4)_3$

$\%=\frac{53.96 g/mol}{342 g/mol}\times 100=15.77\%\approx 15.8\%$

Hence, the correct answer is option (2).

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If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitr

victus00 [196]

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

$3NO_2+H_2O\rightarrow 2HNO_3+NO$

Moles of nitrogen dioxide :

$\frac{25.0 g}{56 g/mol}=0.5434 mol$

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

$\frac{2}{3}\times 0.5434 mol=0.3623 mol$ of nitric acid.

Mass of 0.3623 moles of nitric acid :

$0.3623 mol\times 63 g/mol=22.8260 g$

Theoretical yield = 22.8260 g

Experimental yield = ?

$\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100$

$75\%=\frac{\text{Experimental yield}}{22.8260 g}$

Experimental yield of nitric acid = 17.1195 g

7 0

4 years ago

The dissociation of calcium carbonate has an equilibrium constant of Kp= 1.20 at 800°C. CaCO3(s) ⇋ CaO(s) + CO2(g)

Fed [463]

Explanation:

(a) Formula that shows relation between $K_{c}$ and $K_{p}$ is as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

Here, $\Delta n$ = 1

Putting the given values into the above formula as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

= $1.20 \times (RT)^{-1}$

= $\frac{1.20}{0.0820 \times 1073}$

= 0.01316

(b) As the given reaction equation is as follows.

$CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)$

As there is only one gas so ,

$p[CO_{2}] = K_{p}$ = 1.20

Therefore, pressure of $CO_{2}$ in the container is 1.20.

(c) Now, expression for $K_{c}$ for the given reaction equation is as follows.

$K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}$

= $\frac{x \times x}{(a - x)}$

= \frac{x^{2}}{(a - x)}[/tex]

where, a = initial conc. of $CaCO_{3}$

= $\frac{22.5}{100} \times 9.56$

= 0.023 M

0.0131 = $\frac{x^{2}}{0.023 - x}$

x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

% of $CaCO_{3}$ remained = $(\frac{0.017}{0.023}) \times 100$

= 75.46%

Thus, the percentage of calcium carbonate remained is 75.46%.

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4 years ago

Examine the statement.

sweet [91]

This is an exothermic

8 0

3 years ago

Example of a neutral mutation?

Norma-Jean [14]

The answer is D

Explanation: A RED FLOWER WITH ONE PART COLORED YELLOW

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3 years ago

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At what atom of the pyridine ring will oxidation and reduction reactions take place?a. 2.b. 4.c. 3.d. 1.

bagirrra123 [75]

Answer:

Explanation:

Pyridine is a six membered hetrocyclic aromatic compound. Oxidation of the pyridine ring occurs at position-1.

Just like benzene, pyridine is stable towards mild oxidizing agents, however, pyridine is oxidized by peracetic acid to yield pyridine N-oxide.

7 0

3 years ago