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netineya [11]
3 years ago
10

Given the following reaction: 2D(g) + 3E(g) + F(g) \longrightarrow⟶ 2G(g) + H(g) When the concentration of D is decreasing by 0.

10 M/s, how fast is the concentration of H increasing? When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing? What is the rate of reaction?
Chemistry
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

E decreseas 3/2 as fast as G increases = 0.30 M/s

Explanation:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:

Given data = d[D]/dt = 0.10 M/s

-d[D] / 2dt  = d[H]/dt

d[H]/dt = 0.05 M/s

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:

d[G] / 2dt  = -d[H]/3dt

E decreseas 3/2 as fast as G increases = 0.30 M/s

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During a synthesis reaction, 1.8 grams of magnesium reacted with 6.0 grams of oxygen. What is the maximum amount of magnesium ox
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2.9 grams.

Explanation:

  • From the balanced reaction:

<em>Mg + 1/2O₂ → MgO,</em>

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

  • We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:

no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.

no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.

<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>

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<em><u>Using cross multiplication:</u></em>

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.074 mol of Mg produce → 0.074 mol of MgO.

<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>

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1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

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(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

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