In the following chemical reaction between H_2 and Cl_2 to produce HCl, what is the mass of HCl produced and leftover reactants when 0.36 g of H_2 completely reacts with 12.41 g of Cl_2?I

The total amount of leftover reactants and HCl is 12.79 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of hydrogen gas = 0.36 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

Given mass of chlorine gas = 12.41 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of hydrogen gas and chlorine gas is:

By Stoichiometry of the reaction:

1 moles of chlorine gas reacts with 1 mole of hydrogen gas

So, 0.175 moles of chlorine gas will react with = of hydrogen gas

As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, chlorine gas is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (hydrogen gas) = [0.18 - 0.175] = 0.005 moles

By Stoichiometry of the reaction

1 moles of chlorine gas produces 2 moles of HCl

So, 0.175 moles of chlorine gas will produce = moles of HCl

Now, calculating the mass of hydrogen gas left and HCl from equation 1, we get:

Molar mass of hydrogen gas = 2 g/mol

Moles of excess hydrogen gas = 0.005 moles

Putting values in equation 1, we get:

Molar mass of HCl = 36.5 g/mol

Moles of HCl = 0.350 moles

Putting values in equation 1, we get:

Total mass of HCl and leftover reactants = [12.78 + 0.01] = 12.79 g

Hence, the total amount of leftover reactants and HCl is 12.79 grams

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

the dependant variable is spilled water

the independent is the heat