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3 years ago

34g aluminum are combined with 39g chlorine gas which is the limiting reactant

Chemistry

1 answer:

vampirchik [111]3 years ago

6 0

The limiting reactant is chlorine (Cl2).

Explanation:

Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.

2 Al + 3 Cl2 ==> 2 AlCl3 represents the balanced equation.

Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g

= 1.260 g moles of Al

Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g

= 1.10 g moles of Cl2

Dividing each reactant by it's coefficient in the balanced equation obtains:

1.260 moles Al / 2 = 0.63 g moles of Al

1.11 moles Cl2 / 3 = 0.36 g moles of Cl2

The reactant which produces a lesser amount of product is called as limiting reactant.

Here the Limiting reactant is Cl2.

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Irina-Kira [14]

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

$m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}$

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:

$1 = \%_{16} + \%_{17} + \%_{18}$

$1 = x + 3.2 \cdot 10^{-4} + \%_{18}$

$\%_{18} = 1 - x - 3.2 \cdot 10^{-4}$

Hence, the percent abundance of O-18 is:

$m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}$

$15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)$

$x = 0.980614 \times 100 = 98.0614 \%$

Hence, the percent abundance of oxygen-18 is:

$\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%$

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!

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