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3 years ago

Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.

Chemistry

2 answers:

sergejj [24]3 years ago

5 0

The molecular formula of compound is $Al(C_{2}H_{3}O_{2})_{3}$ .

Since, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6.023\times 10^{23}$ atoms of $Al(C_{2}H_{3}O_{2})_{3}$ .

Thus, according to molecular formula, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6\times 6.023\times 10^{23}=3.61\times 10^{24}$ atoms of oxygen atoms.

1 atom of oxygen will be present in $\frac{1}{3.61\times 10^{24}}$ moles of $Al(C_{2}H_{3}O_{2})_{3}$ . Thus,

$2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}$ .

Molar mass of $Al(C_{2}H_{3}O_{2})_{3}$ is 236 g/mol, mass can be calculated as follows:

$m=n\times M=0.7285 mol\times 236 g/mol=172 g$

Therefore, mass of $Al(C_{2}H_{3}O_{2})_{3}$ will be 172 g.

Vesnalui [34]3 years ago

5 0

Answer : The mass of $Al(C_2H_3O_2)_3$ is, 148.48 grams.

Solution : Given,

Molar mass of $Al(C_2H_3O_2)_3$ = 204 g/mole

As we know that,

1 mole contains $6.022\times 10^{23}$ number of atoms

In the given compound $Al(C_2H_3O_2)_3$ , there are 1 atom of aluminium, 6 atoms of carbon, 9 atoms of hydrogen and 6 atoms of oxygen.

As, $6\times (6.022\times 10^{23})$ number of atoms of oxygen present in 1 mole of $Al(C_2H_3O_2)_3$

So, $2.63\times 10^{24}$ number of atoms of oxygen present in $\frac{2.63\times 10^{24}}{6\times (6.022\times 10^{23})}=0.727$ mole of $Al(C_2H_3O_2)_3$

Now we have to calculate the mass of $Al(C_2H_3O_2)_3$ .

Formula used :

$\text{Mass of }Al(C_2H_3O_2)_3=\text{Moles of }Al(C_2H_3O_2)_3\times \text{Molar mass of }Al(C_2H_3O_2)_3$

Now put all the given values in this formula, we get

$\text{Mass of }Al(C_2H_3O_2)_3=0.727mole\times 204g/mole=148.48g$

Therefore, the mass of $Al(C_2H_3O_2)_3$ is, 148.48 grams.

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Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

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$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

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$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

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$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

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