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gavmur [86]
3 years ago
5

A 1.00 L volume of HCl reacted completely with 2.00 L of 1.50 M Ca(OH)2 according to the balanced chemical equation below. 2HCl

+ Ca(OH)2 > CaCl2 + 2H2O What was the molarity of the HCl solution?
it is either 6.00 M or 0.375 M
Chemistry
2 answers:
kramer3 years ago
6 0

Answer:

D)- 6.00M

Explanation:

took test on edg

Zigmanuir [339]3 years ago
5 0
The first step is to find the number of moles of OH⁻ that reacted with the HCl.  To do this multiply 2.00L by 1.50M to get 3 moles of Ca(OH)₂.  Then you multiply 3 by 2 (there are 2 moles of OH⁻ per every 1 mole of Ca(OH)₂) to get 6 moles of OH⁻.  That means that you needed 6 moles of HCl since 1 mole of HCl contains 1 mole of H⁺ and equal amounts H⁺ and OH⁻ reacted with each other.  To find the molarity of the HCl solution you need to divide 6mol by 1L to get 6M.  Tat means that the concentration of the acid was 6M.

I hope this helps.  Let me know if anything was unclear.
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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
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Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
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