Chemistry

In the reaction: P4 (S) +Cl2 (g)PCl5(s),
1. If you have 10.00 grams of Phosphorus to react with 220 grams of Chlorine
gas, what would be the theoretical yield of PCl5(s), in grams?

2. Why was there condensation of the beaker of cold water when a blue flame
was passes underneath?
(A) the product of carbon dioxide from combustion
(B) the product of hydrogen dioxide from combustion
(C) left over unburned oxygen
(D) left over unburned methane
(E) melting of the glass

3. The reaction of hydrogen peroxide in the presence of manganese dioxide produces:
(A) carbon dioxide
(B) oxygen
(C) hydrogen gas
(D) nitrogen dioxide
(E) methane

4. A 21.18 mL of 0.250 M NaOH is titrated with a H2SO4 solution. The initial volume of H2SO4 was 13.28 and the final volume of H2SO4 was 28.29 mL when the solution turned very slightly pink. What is the concentration of Ca(OH)2?

5. You know the solution forms a precipitate when:
(A) when it changes color
(B) when you can’t see through it
(C) because it bubbles
(D) you have stirred it for more than 5 minutes
(E) the solution thickens

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asked by Jerome
  1. 1. Write the equation and balance it.
    P4 + 10Cl2 ==> 4PCl5

    2a. Convert grams P to moles. mole = grams/molar mass
    2b. Convert g Cl2 to mole. Same process.

    3a. Using the coefficients in the balanced equation, convert moles P to moles product.
    3b. Same procedure, convert moles Cl2 to moles product.
    3c. It is likely that moles product from 3a and 3b will not agree which means one of them is wrong. The correct value, in limiting reagent problems, is ALWAYS the smaller one and the reagent producing that value is the limiting regent.

    4. Now use the smaller value from 3c and convert to grams. g = moles x molar mass. This is the theoretical yield.

    #2. I have no idea what you are talking about in #2.
    #3.oxygen
    #4. You need to read #3 and make changes. M Ca(OH)2 is zero the way I see it; you didn't start with any.
    #5. I don't know what you are doing in the experiment.
    #4.

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    posted by DrBob222

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