Chemistry

can you please please help me with these questions I don't know how to do any of them. I know how to calculate moles and grams and mole ratio,etc but I don't understand the limiting excess reactant and yield stuff. Thanks!


1.In the following chemical reaction, how many grams of the excess reactant are left over when 0.45 moles of Al react with 1.25 moles of Cl2 ?
2Al + 3Cl2 → 2AlCl3


2.Using the following equation, determine the % yield from the following reaction if 30.65 g of octane (C8H18 ) react with excess oxygen to produce 81.75 g of CO2

C8 H18 (I) + 25 O2 (g) → 16CO2 + 18H2O

moles of kcl=10.35/74.55=0.138


3.
For the following reaction, determine the amount of AgCl(s) formed when 10.35 g of KCl(aq) react with excess AgNO3 (aq) to produce AgCl(s) and KNO3 (aq) with a 77.15% yield.

AgNO3 + KCl → AgCl(s) + KNO3

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asked by molly
  1. 1.In the following chemical reaction, how many grams of the excess reactant are left over when 0.45 moles of Al react with 1.25 moles of Cl2 ?
    2Al + 3Cl2 → 2AlCl3

    LR = limiting reagent. ER - excess reagent. Actually you work three regular stoichiometry problems here. First you want to identify the LR and ER and you do it this way.Use the coefficients in the balanced equation to convert mols Al AND mols Cl2 (in two separate operations) to mols AlCl3.
    0.45 mols Al x (2 mols AlCl3/2 mols Al ) = 0.45 mols AlCl3 formed.

    1.25 mols Cl2 x (2 mols AlCl3/3 mols Cl2) = 0.83 mols AlCl3.
    IN LR problems the smaller number is ALWAYS the mols Product formed and the reagent responsible for that is the LR. In this case that makes Al the LR and Cl2 the ER.
    Now that you know Cl2 is the ER, then you need to calculate the amount of Cl2 used. To do that just do the stoichiometry again; i.e., 0.45 mols Al x (3 mols Cl2/2 mols Al) = 0.68 mols Cl2 used.
    1.25 mols Cl2 initially.
    -0.68 mols Cl2 used
    --------
    1.25-0.68 = ? mols Cl2 not reacted.


    2.Using the following equation, determine the % yield from the following reaction if 30.65 g of octane (C8H18 ) react with excess oxygen to produce 81.75 g of CO2

    C8 H18 (I) + 25 O2 (g) → 16CO2 + 18H2O

    moles of kcl=10.35/74.55=0.138
    I don't know where this 0.138 mols KCl fit so I'll ignore it.

    For problem #2, you say you know how to vind mols etc. Do this.
    a. Convert grams octane to mols.
    b. Use the coefficients in the balanced equation to convert mols octane to mols CO2.
    c. Convert mols CO2 to grams CO2. THIS IS THE THEORETICAL YIELD (THE AMOUNT OF STUFF FORMED FOR 100% YIELD). I call this TY for theoretical yield. The actual yield (AY) is 81.75 g
    d. %yield = (AY/TY)*100 = ?


    3.
    For the following reaction, determine the amount of AgCl(s) formed when 10.35 g of KCl(aq) react with excess AgNO3 (aq) to produce AgCl(s) and KNO3 (aq) with a 77.15% yield.

    AgNO3 + KCl → AgCl(s) + KNO3

    This looks like #2.

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    posted by DrBob222
  2. thank you so much that makes so much sense :)

    1. 0
    2. 0
    posted by molly

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