manganese iv oxide reacts with HCI to form MnO2, Cl2 has, and H2O determine the amount of each chemical present at the end of the reaction if 74.8 grams of MnO2 reacts with 48.2 grams of HCI

Question

please help, i really dont understand this

Answer 1

First, let us write out the balanced chemical equation:

MnO₂ +4HCl → MnCl₂ + Cl₂ + 2H₂O

We must then find the limiting reagent. Let us test 74.8 g MnO₂ first:

74.8 g MnO₂ * (1 mol MnO₂/86.9386 g MnO₂) * (4 mol HCl/1 mol MnO₂) * (36.46 g HCl/1 mol HCl) =

125.48 g HCl

Immediately, we see that 74.8 g of MnO₂ will react with 125.48 g of HCl. Since we know we have less grams of HCl than that (48.2 g), we know that HCl is our limiting reagent. Thus, we can determine the amount of all the products using our initial mass of HCl:

48.2 g HCl * (1 mol HCl/36.46 g HCl) * (1 mol MnCl₂/1 mol HCl) * (125.844 g MnCl₂/1 mol MnCl₂) =

166.37 g MnCl₂

48.2 g HCl * (1 mol HCl/36.46 g HCl) * (1 mol Cl₂/1 mol HCl) * (70.906 g Cl₂/1 mol Cl₂) =

93.74 g Cl₂

48.2 g HCl * (1 mol HCl/36.46 g HCl) * (2 mol H₂O/1 mol HCl) * (18 g H₂O/1 mol H₂O) =

47.59 g H₂O

Answer 2

This is a limiting reagent problem. We are performing a reaction and want to know how much of each substance we have at the end of it. More than likely we have too much of one of the reactants (HCl or MnO₂) so we will have some left over. Solve limiting reagent problems systematically.

1) Figure out the balanced equation.

They tell us we are reacting MnO₂ with HCl to form MnCl₂ + H₂O and Cl₂. If we write that and balance it we have

MnO₂ + 4HCl -> Cl₂ + 2H₂O + MnCl₂

2) Convert what you are given into moles.

We always work in moles when we do reactions, so let's convert our known masses to moles.

74.8g MnO₂ = 0.86 moles MnO₂ (divide the 74.8g / 86.94g/mol to get mol)

48.2g HCl = 1.32 moles HCl (same math as above using HCl molar mass)

3) Find the limiting reactant.

This is usually the hard part. When we find the limiting reagent we are seeking the answer to 2 questions: 1) What reactant am I going to run out of first

2) What reactant am I going to have left over when the reaction is finished

So let's start with HCl. We could start with MnO₂ but we will arrive at the same answer so I usually just choose randomly. We have 1.32 moles of HCl. How much MnO₂ would it take to react COMPLETELY? We can see from the balanced equation that 1 mole of MnO₂ reacts with 4 moles of HCl (see the coefficients). This is just a ratio! Think of it as 1:4 MnO₂:HCl if that helps.

But we don't have a round number like 1 or 4 moles, we have 1.32 moles of HCl. Using some math we can find out how many moles of MnO₂ we would need for that exact amount of HCl. If we set up an algebra problem it looks like this:

1 mol MnO₂ / 4 mol HCl = x mol MnO₂ / 1.32 mol HCl (Write this out on your own paper with numerators and denominators so it makes more sense)

We we want to solve for x then we have to multiply the left side of the equation by 1.32 mol HCl. If we do this and solve for x we get 0.33 mol MnO₂ required. Now look at how much MnO₂ we actually have. 0.86 mol MnO₂ is more than enough to react with all 1.32 mol of HCl. What does this tell us? It means that We will have MnO₂ left over after the reaction is done which means it is the excess reactant. Thus HCl is the limiting reactant - we will run out of it first.

4) Do the reaction using the limiting reactant.

Here all we are doing is reaction stoichiometry using coefficients from the balanced equation to get how many moles of each product the reaction will produce. The general template for calculating this is:

1.32 mol HCl x (coefficient of product in balanced equation) / 4 mol HCl = moles of product produced

For example using Cl₂ :

1.32 mol HCl x 1 mol Cl₂ / 4 mol HCl = 0.33 mol Cl₂ produced. Continuing on in the same manner for the other two products we should get 0.66 mol H₂O and 0.33 mol MnCl₂.

5) Find how much excess reactant you have left.

Remember how we said we would only need 0.33 moles of MnO₂ to react with all 1.32 moles of our HCl? At this point we have done that and used up 0.33 moles of MnO₂. If we subtract this amount from the amount of MnO₂ we had at the beginning, that's how much excess reactant we have left. If we do this it's 0.53 moles MnO₂ remaining.

That was a long explanation so take time to read each part carefully and do the work yourself to ensure you understand. Remember at the end to convert each amount to grams if the question asks you to. Feel free to reach out with any questions.