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Allisa [31]
3 years ago
7

At 311 K, this reaction has a K c value of 0.0111 . X ( g ) + 2 Y ( g ) − ⇀ ↽ − 2 Z ( g ) Calculate K p at 311 K. Note that the

pressure is in units of atmosphere (atm).
Chemistry
1 answer:
Aleks [24]3 years ago
4 0

Answer:

K_{p}=4.35\times 10^{-4}

Explanation:

We know, K_{p}=K_{c}(RT)^{\Delta n}

where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and \Delta n is difference in sum of stoichiometric coefficient of products and reactants

Here \Delta n=(2)-(2+1)=-1 and T = 311 K

So, K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}

Hence value of equilibrium constant in terms of partial pressure (K_{p}) is 4.35\times 10^{-4}

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Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 6.9 g
satela [25.4K]

Answer:

5.6gNa_2SO_4

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4

Best regards.

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