2 Fe (s) + 6 HCl (aq) -> 2 FeCL3 (s) + 3 H2 (g)

Given 16 g of HCl and 10g Fe react with how many grams of FeCl3 are produced (limiting reactant)?

This is a limiting reagent problem. Here is a worked example.

http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html

thanks a bunch!

To determine the limiting reactant and calculate the mass of FeCl3 produced, we need to compare the amount of product that would be formed from each reactant.

1. Calculate the number of moles for each reactant:
- HCl: 16 g / molar mass of HCl
- Fe: 10 g / molar mass of Fe

2. Use the balanced equation to find the stoichiometric ratio between the reactants and the product:
From the balanced equation: 2 Fe (s) + 6 HCl (aq) -> 2 FeCl3 (s) + 3 H2 (g)
The stoichiometric ratio between Fe and FeCl3 is 2:2, meaning that for every 2 moles of Fe reacted, 2 moles of FeCl3 will be produced.

3. Calculate the amount of FeCl3 produced from each reactant:
- Fe: Moles of Fe * Stoichiometric ratio of FeCl3/Fe
- HCl: Moles of HCl * Stoichiometric ratio of FeCl3/HCl

4. Compare the amounts of FeCl3 produced from each reactant:
- The reactant that produces the lesser amount of FeCl3 is the limiting reactant, as it will determine the maximum amount of product that can be formed.

5. Calculate the mass of FeCl3 produced from the limiting reactant:
- Use the number of moles of FeCl3 produced from the limiting reactant and multiply it by the molar mass of FeCl3 to obtain the mass.

Let's go through the calculations step-by-step:

1. Calculate the number of moles for each reactant:
- HCl: 16 g / 36.46 g/mol = 0.4398 mol
- Fe: 10 g / 55.85g/mol = 0.179 mol

2. Use the balanced equation to find the stoichiometric ratio:
2 Fe (s) + 6 HCl (aq) -> 2 FeCl3 (s) + 3 H2 (g)

3. Calculate the amount of FeCl3 produced from each reactant:
- Fe: 0.179 mol * (2 mol FeCl3 / 2 mol Fe) = 0.179 mol FeCl3
- HCl: 0.4398 mol * (2 mol FeCl3 / 6 mol HCl) = 0.1466 mol FeCl3

4. Compare the amounts of FeCl3 produced from each reactant:
- The limiting reactant is HCl because it produces a lesser amount of FeCl3 (0.1466 mol).

5. Calculate the mass of FeCl3 produced from the limiting reactant:
- Mass of FeCl3 = Moles of FeCl3 * Molar mass of FeCl3
- Mass of FeCl3 = 0.1466 mol * (162.204 g/mol) = 23.8 g

Therefore, the mass of FeCl3 produced (limiting reactant) is 23.8 grams.

To determine the limiting reactant and calculate the amount of FeCl3 produced, we need to compare the amount of product that can be formed from each reactant.

First, we need to convert the given masses of HCl and Fe to moles. To do this, we need to know the molar mass of each substance.

The molar mass of HCl is:
H: 1 g/mol
Cl: 35.5 g/mol
So, the molar mass of HCl is 1 + 35.5 = 36.5 g/mol.

Next, we can calculate the number of moles of HCl present:
Number of moles = mass / molar mass
Number of moles of HCl = 16 g / 36.5 g/mol = 0.44 mol

Similarly, we can calculate the number of moles of Fe:
The molar mass of Fe is:
Fe: 55.85 g/mol
Number of moles of Fe = 10 g / 55.85 g/mol = 0.18 mol

From the balanced equation, we can see that the molar ratio of Fe to FeCl3 is 2:2, which means that 2 moles of Fe react to form 2 moles of FeCl3. Therefore, the stoichiometric ratio of Fe to FeCl3 is 1:1.

Now, compare the number of moles of Fe and HCl. We have 0.18 moles of Fe and 0.44 moles of HCl.

Since the stoichiometric ratio is 1:1, it means that 0.18 moles of Fe will react with 0.18 moles of HCl. This is the smaller value of moles between the two.

Now, we can calculate the mass of FeCl3 produced from 0.18 moles of FeCl3:
Mass of FeCl3 = moles of FeCl3 × molar mass of FeCl3

The molar mass of FeCl3 is:
Fe: 55.85 g/mol
Cl: 35.5 g/mol (x3, since there are three chlorine atoms)
Molar mass of FeCl3 = 55.85 + (35.5 × 3) = 162.35 g/mol

Mass of FeCl3 = 0.18 mol × 162.35 g/mol = 29.22 g

Therefore, when 10g of Fe and 16g of HCl react, the limiting reactant is Fe, and the mass of FeCl3 produced is 29.22 g.