Hydrogen and chlorine react to yield hydrogen chloride: H2+ Cl2 ( 2HCl. How many grams of HCl are formed from reaction of 3.56 g of H2 with 8.94 g of Cl2? Which reactant is limiting?

I do limiting reagent problems the long way. There are shorter ways to work this problem but I think they are more difficult to explain.

H2 + Cl2 ==> 2HCl

mols H2 = grams/molar mass = 3.56/2 = 1.78
mols Cl2 = 8.94/71 = 0.126

Using the coefficients in the balanced equation, we now convert each to mols HCl produced if we had all of the other compound we needed.
1.78 mols H2 x (2 mol HCl/1 mol H2) = 1.78 x 2/1 = 3.56 mol HCl.

0.126 mols Cl2 x (2 mol HCl/1 mol Cl2) = 0.126 x 2/1 = 0.252 mol HCl.

The two values HCl produced do not agree; in limiting reagent problems the correct value is ALWAYS the smaller value; therefore, Cl2 is the limiting reagent and 0.252 mol HCl will be produced.

Convert mols HCl to g HCl.
grams = mols x molar mass = ?

9.2g HCl

Well, let's calculate the number of moles for each reactant.

For hydrogen (H2):
Molar mass of H2 = 2 g/mol
Number of moles = 3.56 g / 2 g/mol = 1.78 mol

For chlorine (Cl2):
Molar mass of Cl2 = 70.91 g/mol
Number of moles = 8.94 g / 70.91 g/mol = 0.126 mol

Now, to determine the limiting reactant, we compare the ratios of the number of moles of each reactant to their coefficients in the balanced equation.

From the balanced equation:
2 moles of H2 reacts with 1 mole of Cl2 to produce 2 moles of HCl.

For hydrogen (H2):
1.78 mol H2 * (2 mol HCl / 2 mol H2) = 1.78 mol HCl

For chlorine (Cl2):
0.126 mol Cl2 * (2 mol HCl / 1 mol Cl2) = 0.252 mol HCl

So, from the calculations, we can see that the chlorine (Cl2) is the limiting reactant.

Now, to calculate the mass of HCl formed from the limiting reactant:
Mass = number of moles * molar mass
Mass of HCl formed = 0.252 mol HCl * 36.46 g/mol = 9.19 g HCl

Therefore, 9.19 grams of HCl are formed from the reaction of 3.56 g of H2 with 8.94 g of Cl2.

To determine the number of grams of HCl formed and the limiting reactant, we need to use stoichiometry and compare the number of moles of each reactant.

First, we need to convert the given masses of H2 and Cl2 into moles. We can use the molar masses to do this:

Molar mass of H2 = 2 g/mol
Molar mass of Cl2 = 71 g/mol

Moles of H2 = mass of H2 / molar mass of H2
= 3.56 g / 2 g/mol
= 1.78 mol

Moles of Cl2 = mass of Cl2 / molar mass of Cl2
= 8.94 g / 71 g/mol
= 0.126 mol

Now let's determine the stoichiometry of the reaction:

2H2 + Cl2 → 2HCl

From the balanced equation, we can see that the ratio of moles of H2 to moles of HCl is 2:2, or 1:1. This means that for every 1 mole of H2, we will get 1 mole of HCl.

Since we have 1.78 moles of H2, we will generate 1.78 moles of HCl.

Now let's calculate the mass of HCl formed:

Mass of HCl = moles of HCl * molar mass of HCl
= 1.78 mol * 36.5 g/mol (molar mass of HCl)
= 64.87 g

Therefore, 64.87 grams of HCl are formed.

To determine the limiting reactant, we need to compare the number of moles of each reactant used in the reaction.

From the calculations above, we have:
Moles of H2 = 1.78 mol
Moles of Cl2 = 0.126 mol

The reactant with the smallest number of moles is the limiting reactant. In this case, Cl2 has the smallest number of moles (0.126 mol). Therefore, Cl2 is the limiting reactant.

To find out how many grams of HCl are formed from the reaction, you need to use stoichiometry and the concept of limiting reactants.

1. Write the balanced equation for the reaction:
H2 + Cl2 -> 2HCl

2. Calculate the moles of each reactant:
Molar mass of H2 = 2.02 g/mol
Molar mass of Cl2 = 70.9 g/mol

Moles of H2 = mass / molar mass = 3.56 g / 2.02 g/mol = 1.76 mol
Moles of Cl2 = mass / molar mass = 8.94 g / 70.9 g/mol = 0.126 mol

3. Determine the limiting reactant:
To find the limiting reactant, compare the mole ratio of reactants in the balanced equation to the actual mole ratio you calculated from the amounts given.

From the balanced equation, you can see that 1 mole of H2 reacts with 1 mole of Cl2 to produce 2 moles of HCl.

The mole ratio between H2 and Cl2 is 1:1.76 (approximately 1:2) based on the mass given.
The mole ratio between H2 and Cl2 in the balanced equation is 1:0.126 (approximately 1:0.13) based on stoichiometry.

Since the mole ratio of H2 to Cl2 is lower in stoichiometry compared to the actual ratio calculated from the masses, H2 is the limiting reactant.

4. Calculate the moles of HCl formed:
From the balanced equation, you can see that 2 moles of HCl are formed for every mole of H2.

Moles of HCl = moles of H2 x (2 moles of HCl / 1 mole of H2) = 1.76 mol x 2 = 3.52 mol

5. Calculate the mass of HCl formed:
Molar mass of HCl = 36.5 g/mol

Mass of HCl = moles of HCl x molar mass = 3.52 mol x 36.5 g/mol = 128.28 g

Therefore, 128.28 grams of HCl are formed from the reaction between 3.56 g of H2 and 8.94 g of Cl2. The limiting reactant is H2.