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Tanya [424]
3 years ago
5

Calculate the concentration in mol L^-1 of ammonium ions in a solution prepared by mixing 360 mL of 0.250 mol L^-1 ammonium sulf

ate solution with 675.0 mL of 1.20 mol L^-1 ammonium nitrate solution. Assume solutions are additive.
Chemistry
1 answer:
Mnenie [13.5K]3 years ago
8 0

In order to determine the concentration of ammonium ions in the solution prepared by mixing solutions of ammonium sulfate, (NH4)2SO4, and ammonium nitrate, first calculate the amount of ammonium ions for each solution.<span>

<span>For ammonium sulfate sol'n: 0.360 L x 0.250 mol(NH4)2SO4/ L x 2 mol NH4+ /1 mol(NH4)2SO4 = 0.18 mol NH4+
<span>For ammonium nitrate sol'n: 0.675 x 1.2 mol NH4NO3/L x 1 mol NH4+ /1 molNH4NO3 = 0.81 mol NH4+

Thus, the amount of NH4+ ions is (0.18 + 0.81) mol or 0.99 mol NH4+. To get the concentration, multiply this to the volume of solution which is assumed to be additive, such that:</span></span></span>

M NH4+ in sol’n = 0.99 mol NH4+/1.035 L = 0.9565 mol NH4+/ L sol’n

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3 years ago
Consider the reaction: P4 + 6Cl2 = 4PCl3.
likoan [24]

Answer:

The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g___________

                                P4      +      6Cl2      =      4PCl3

                          4(31) ---------- 12(35.5)

                         20     ----------    x

                    x = 20(12x35.5) / 4(31)

                   x = 8520 / 124

                   x = 68.7 g

b. You have 15.00 g. of P4 and 22.00 g. of Cl2, identify the limiting reactant and calculate the grams of PCl3 that can be produced as well as the grams of excess reactant remaining. LR____________ grams PCl3 _________ grams excess reactant ___________

                            P4      +      6Cl2      =      4PCl3

                       124g             426 g               4(31 + 3(35.5)) = 550g

                        15g               22g

I will use P4 to find the limiting reactant

                 

                     x = (15 x 426) / 124 = 51.5   The limiting reactant is Chlorine

                                                                  because we need 51.5 g and we only have 22g

Excess reactant

                 x = (22 x 124) / 426 = 6.4 g of P4

           Excess P4 = 15 g - 6.4 = 8.6 g of P4 in excess

Grams of PCl3 produced

                              426 g of Cl2 ----------------  550 g of PCl3

                                 22g of Cl2 ------------- -     x

            x = (22 x 550) / 426 = 28.4 g of PCl3

c. If the actual amount of PCl3 recovered is 16.25 g., what is the percent yield? ______________

   % yield = (16.25  - 28.4) / 28.4 x 100

  % yield = 42.8

d. Given 28.00 g. of P4 and 106.30 g. of Cl2, identify the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______________ grams excess reactant

Limiting reactant

                                   124g of P4  -------------      426 g  6Cl2

                                     28g           ---------------     x

x = (28 x 426) / 124

x = 96.2 g of Cl2 and we have 106.3 so Chlorine is the excess reactant and P4 is the limiting reactant.

Excess reactant = 106.3  - 96.2 = 10.1 g of Cl2 in excess

                   

                 

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<h3>What is  a Lewis acid?</h3>

A Lewis acid is a substance that accepts electron pair while a Lewis base donates an electron pair.

Now consider the given reaction; AlCl3 +Cl^- ------> AlCl 4 ^-. The Lewis acid here is AlCl3  while the Lewis base here is Cl^- .

Learn more about acid - base reaction: brainly.com/question/14356798

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How many moles of sodium ions are present in 0.25 kg of a 2.0 m solution of sodium chloride in water
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