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Kay [80]
2 years ago
6

If a chlor-alkali cell used a current of 3X10⁴A, how many pounds of Cl₂ would be produced in a typical 8-h operating day?

Chemistry
1 answer:
vova2212 [387]2 years ago
5 0

7 × 10² pounds of Cl₂ would be produced in a typical 8-h operating day.

<h3>What is Stoichiometry ?</h3>

Stoichiometry helps us use the balanced chemical equation to measures quantitative relationships and it is to calculate the amount of products and reactants that are given in a reaction.

<h3>What is Balanced chemical equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactants side is equal to the number of atoms on the product side in an equation.

2Cl⁻ (aq) → Cl (g) + 2e⁻

According to stoichiometry, moles of Cl₂

= (3 \times 10^4\ A) \left (\frac{\frac{C}{s}}{A} \right ) \left (\frac{3600\ s}{h} \right ) (8h) \left (\frac{1\ \text{mol}\ e^{-}}{9.65 \times 10^4\ C} \right ) \left (\frac{1\ \text{mol}\ Cl_2}{2\ \text{mol}\ e^-} \right )

= 4477 moles

Pounds Cl₂

= (4477\ \text{moles}\ Cl_2) \left (\frac{70.90\ g\ Cl_2}{\text{mol}} \right ) \left (\frac{1\ kg}{1000\ g} \right ) \left (\frac{2.205\ lb}{1\ kg} \right )

= 7 × 10² lb

Thus from the above conclusion we can say that 7 × 10² pounds of Cl₂ would be produced in a typical 8-h operating day.

Learn more about Stoichiometry here: brainly.com/question/14935523

#SPJ4

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A molecule of pentane is made of 5 carbon atoms and 12 hydrogen atoms. a collection of pentane molecules has 3.01 x 1024 carbon
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3.01 Ă— 10^24 Ă— (12/5) hydrogen atoms  
Looking at the formula for the molecule, the ratio of carbon to hydrogen atoms is 5:12, so if we divide the number of carbon atoms by 5 and then multiply by 12, we can find the number of hydrogen atoms. Let's look at the available options and see what makes sense.  
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When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsiu
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Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

<em>1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄</em>

<em />

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

<em>Moles S₈: 256.52g/mol</em>

10.0g * (1mol / 256.52g) = 0.0390 moles

<em>Moles F₂:</em>

<em>PV = nRT</em>

PV/RT = n

<em>Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)</em>

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

0.600 moles = n

For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

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