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3 years ago

How many moles of PC15 can be produced from 51.0 g of Cl2 (and excess P4)?

Chemistry

1 answer:

slava [35]3 years ago

8 0

Answer:

0.287 mole of PCl5.

Explanation:

We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 = 51g

Number of mole of Cl2 =..?

Mole = Mass /Molar Mass

Number of mole of Cl2 = 51/71 = 0.718 mole

Next, we shall write the balanced equation for the reaction. This is given below:

P4 + 10Cl2 → 4PCl5

Finally, we determine the number of mole of PCl5 produced from the reaction as follow:

From the balanced equation above,

10 moles of Cl2 reacted to produce 4 moles of PCl5.

Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.

Therefore, 0.287 mole of PCl5 is produced from the reaction.

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Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

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Which place in the world has the highest average annual precipitation?

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The correct name for hno2 is ________. nitrous acid hyponitrous acid nitric acid pernitric acid hydrogen nitrate

balu736 [363]

Nitrous Acid.

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Nitric acid: HNO3

Pernitric acid: HNO

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Sort these species into isoelectronic groups. It doesn\'t matter which group goes in which box, so long as the correct species a

OleMash [197]

Answer : The isoelectronic groups are:

$He,Be^{2+},Li^{+}$

$N^{3-},Mg^{2+}$

$S^{2-},Ti^{4+},K^{+}$

Explanation :

Isoelectronic : It is defined as the compound or molecule having the same number of electrons and the same number of electronic structure.

The element is helium. The number of electrons are 2.
The element is beryllium. The number of electrons are 4. The number of electrons in $Be^{2+}$ = 4 - 2 = 2
The element is lithium. The number of electrons are 3. The number of electrons in $Li^{+}$ = 3 - 1 = 2
The element is nitrogen. The number of electrons are 7. The number of electrons in $N^{3-}$ = 7 + 3 = 10
The element is neon. The number of electrons are 10.
The element is sulfur. The number of electrons are 16. The number of electrons in $S^{2-}$ = 16 + 2 = 18
The element is magnesium. The number of electrons are 12. The number of electrons in $Mg^{2+}$ = 12 - 2 = 10
The element is titanium. The number of electrons are 22. The number of electrons in $Ti^{4+}$ = 22 - 4 = 18
The element is potassium. The number of electrons are 19. The number of electrons in $K^{+}$ = 19 - 1 = 18