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babymother [125]
3 years ago
14

I. Determine the mass of reactants and products that will be needed if one mole of P4(s) reacts completely, as follows

Chemistry
1 answer:
natka813 [3]3 years ago
7 0

Answer:

I. <em>Reagents: </em>

P₄= 123.88 g

H₂ = 12 g

<em>Products: </em>

PH₃= 135.88g

II. Total reagents = 135.88 g

Total products = 135.88 g

III. The<em> principle of conservation of mass</em>

Explanation:

I. First, the molar masses of all reagents and all reaction products are calculated:

• mP₄ = 4 x mP = 4 x 30.97g = <em>123.88 g / mol </em>

• mH₂ = 2 x mH = 2 x 1.00g = <em>2 g / mol </em>

• mPH₃ = 4 x mH + m P = 3 x 1.00g + 30.97g = <em>33.97 g / mol </em>

Having the equation balanced, it can be seen that in order <u><em>for one mole to react completely of P₄, 6 moles of H₂ must react, and 4 moles of PH₃ will be produced</em></u>. With the molar masses, we obtain the reacting masses of each reagent and the mass of product that is formed:

<em>Reagents: </em>

P₄: 1 mol ≡ 123.88 g

H₂: 6 moles ≡ 12 g

<em>Products: </em>

PH₃: 4 moles ≡135.88g

II. We add the total mass of the reagents:

<em>Total reagents</em> = mP₄ + mH₂ = 123.88 g + 12 g =<em> 135.88 g </em>

As the reaction product is only PH₃,<em> the total mass of products is </em><em>135.88 g</em>

<em />

III.<em> </em>It is seen that the mass of reagents necessary to produce the reaction is equal to the mass of product obtained. Therefore, the principle illustrated with this example is <u><em>the principle of conservation of mass</em></u>, it says that “<em>In an isolated system, during any ordinary chemical reaction, the total mass in the system remains constant, that is, the mass Consumption of reagents is equal to the mass of the products obtained</em> ”

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A) t = 22.5 min and B) t = 29.94 min

Explanation:

Initial concentration, [A]₀ = 100

Final concentration = 100 -75 = 25

Time = 45 min

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ln[A] − ln[A]₀ = −kt

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ln[25] − ln[100] = - 45k

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Time = ?

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Solving for k;

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B) Zero Order

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Final concentration, [A] = 100 -50 = 50

Time = ?

[A] = [A]₀ − kt

50 = 100 - (1.67)t

-50 = - 1.67t

t = 29.94 min

3 0
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