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kirza4 [7]
3 years ago
5

In a single replacement reaction between 2.57 moles Aluminum and 3.59 moles of Hydrochloric acid, how many moles of Hydrogen can

actually be produced
Chemistry
1 answer:
Alenkasestr [34]3 years ago
5 0

Answer:

1.795 mole of H2.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Al + 6HCl —> 2AlCl3 + 3H2

Step 2:

Determination of the limiting reactant.

From the balanced equation above,

2 moles of Al reacted with 6 moles

Therefore, 2.57 moles of Al will react with = (2.57 x 6)/2 = 7.71 moles of HCl.

From the calculation made above, it will require a higher amount of HCl than what was given to react completely with 2.57 moles of Al. Therefore, HCl is the limiting reactant and Al is the excess reactant.

Step 3:

Determination of the number of mole H2 produced from the reaction.

Here, we shall be using the limiting reactant because it will produce the maximum yield of the reaction since all of it were consumed by the reaction.

The limiting reactant is HCl and the amount of H2 produce can be obtained as follow:

From the balanced equation above,

6 moles of HCl reacted to produce 3 moles of H2.

Therefore, 3.59 moles of HCl will produce = (3.59 x 3)/6 = 1.795 mole of H2.

From the calculations made above, 1.795 mole of H2 is produced from the reaction.

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Which of these elements have the strongest and the weakest forces between its atoms.Give A Reason For Your Answer. Copper,oxygen
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The normal boiling point of a liquid is 282 °C. At what temperature (in
ElenaW [278]

Answer:

The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

ln\left (\frac{p_{2}}{p_{1}}  \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}}  \right )

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

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R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

ln\left (\frac{0.2}{1}  \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15}  \right )

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To convert to Celsius degree temperature, we subtract 273.15 as follows

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Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

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3 years ago
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