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vesna_86 [32]
3 years ago
11

The alkali metals react with the halogens to form ionic metal halides. what mass of potassium chloride forms when 5.11 l of chlo

rine gas at 0.943 atm and 286 k reacts with 29.0 g potassium?
Chemistry
1 answer:
ale4655 [162]3 years ago
4 0
<span>ideal gas law: PV = nRT so .....</span><span> V = PV/(RT) </span>
<span>
Initial number of moles of Cl, n = 0.943*5.11/(0.08206 × 286) mol = 0.2053 moles.
</span><span>
We know the molar mass of K (potassium) = 39.0 g/mol </span>
<span>sooo....
The Initial number of moles of K = 29.0 g/(39.0 g/mol) = 0.7436 moles</span>

<span>Find the balanced equation for the reaction : </span><span>2K + Cl2 → 2KCl </span>
<span>Mole ratio of K:Cl = 2:1 </span>

<span>So after the reaction, the amount of K needed = (0.2053 mol) × 2 = 0.4106 mol which is less than 0.7436 mol </span>
<span>
This means that K is in excess but Cl completely reacts. </span>

<span> So we know the mole ratio is  Cl:KCl = 1 : 2 
</span>
<span>Number of moles of Cl (completely) reacted = 0.2053 mol which means the n</span><span>umber of moles of KCl formed = (0.2053 mol) × 2 = 0.4106 mol </span>

<span>Molar mass of KCl = (39.0 + 35.5) g/mol = 74.5 g/mol </span>
<span>Mass of KCl formed = 0.4106 mol * 74.5 g/mol = 30.6 g</span>
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The equilibrium constant, Kc, for the following reaction is 56.0 at 278 K. 2CH2Cl2(g) CH4(g) + CCl4(g) When a sufficiently large
BabaBlast [244]

Answer:

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Explanation:

<u>1) Equilibrium equation (given):</u>

  • 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)

<u>2) Write the concentration changes when some concentration, A, of CH₂Cl₂ (g) sample is introduced into an evacuated (empty) vessel:</u>

  • 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)

           A - x                 x              x

<u>3) Replace x with the known (found) equilibrium concentraion of CCl₄ (g) of 0.348 M</u>

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          A - 0.3485       0.348       0.348

<u>4) Write the equilibrium constant equation, replace the known values and solve for the unknown (A):</u>

  • Kc = [ CH₄ (g) ] [ CCl₄ (g) ] / [ CH₂Cl₂ (g) ]²

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7 0
3 years ago
Solution A is 0.44 M and reacts with 0.11 M of solution B. Assume that the value of x is 0, the value of y is 1, and r is 1.07 ×
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Answer:

K, the rate constant = 9.73 × 10^(-1)/s

Explanation:

r = K × [A]^x × [B]^y

r = Rate = 1.07 × 10^(-1)/s

K = Rate constant

A and B = Concentration in mol/dm^-3

A = 0.44M

B = 0.11M

x = Order of reaction with respect to A = 0

y = Order of reaction with respect to B = 1

Solving, we get

r/([A]^x × [B]^y) = K

K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727

K = 0.9727

7 0
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mass=relative  formula mass  x  number  of  moles
That  is   32g/mol x 6  moles  =192grams


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