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5. 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.

  1. First, we need to calculate the number of moles for each reactant.

    The molar mass of iron (Fe) is approximately 55.85 g/mol and the molar mass of chlorine gas (Cl2) 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 = 4.53 mol

  2. The balanced chemical equation tells us that 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. To find out if we have enough Cl2 for all the Fe to react, we multiply the moles of Fe by 3/2.

    Required moles of Cl2 for all Fe to react = 4.51 mol * (3/2) = 6.77 mol

  4. We only have 4.53 mol of Cl2, which is less than the required 6.77 mol. Therefore, Cl2 is the limiting reactant.

  5. Since Cl2 is the limiting reactant, Fe is the excess reactant.

So, the answer is Iron.

This problem has been solved

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