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2. If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the excess reactant?2Fe( s ) + 3Cl2( g ) → 2FeCl3( s )IronChlorine gasIron(III) chlorideNone of the above

Question

  1. If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the excess reactant?2Fe( s ) + 3Cl2( g ) → 2FeCl3( s )IronChlorine gasIron(III) chlorideNone of the above
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Solution

To determine the excess reactant, we first need to find out which reactant is limiting. This is done by comparing the amount of product each reactant can produce.

  1. Calculate the number of moles for each reactant:

    The molar mass of Fe (Iron) is approximately 55.85 g/mol and Cl2 (Chlorine gas) is approximately 70.90 g/mol.

    Moles of Fe = 252 g / 55.85 g/mol = 4.51 mol Moles of Cl2 = 321 g / 70.90 g/mol = 4.52 mol

  2. According to the balanced chemical equation, the mole ratio of Fe to Cl2 is 2:3. This means for every 2 moles of Fe, we need 3 moles of Cl2.

  3. Determine the limiting reactant:

    If we multiply the moles of Fe by 3/2, we get the required moles of Cl2 to react with Fe completely.

    Required moles of Cl2 = 4.51 mol * (3/2) = 6.77 mol

    But we only have 4.52 mol of Cl2, which is less than the required amount. So, Cl2 is the limiting reactant.

  4. Therefore, Fe is the excess reactant.

This problem has been solved

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