400 centimetre cube of 0.5mol/decimetre cube hydrochloric acid was reacted with 5 grams of magnesium oxide.
(A) Calculate & identify the Excess & limiting reactants.
(B) What mass of magnesium actually took part in the reaction?
This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants. The following steps will work almost any LR problem.
1. Write and balance the equation.
Mg + 2HCl ==> MgCl2 + H2
2. Convert reactants to mols
a. mols HCl = M x L = 0.5 x 0.400 = approx 0.2
b. mols Mg = grams/atomic mass = 5/24.3 = approx 0.21
3a. Using the coefficients in the balanced equation, convert mols HCl to mols of EITHER product. I'll choose H2.
0.2 mols Mg x (1 mol H2/1 mol Mg) = 0.2 x 1/1 = 0.2 mols H2 (if we used all of it and had all of the HCl needed to react completely).
3b. Do the same and convert mols HCl to mols H2 (use H2 since I used that in 3a).
0.21 mols HCl x (1 mol H2/2 mols HCl) = 0.21 x 1/2 = 0.105 mols H2 produced if we used all of the HCl and had all of the Mg we needed.
3c. You can see that the mols H2 produced is not the same; the correct answer in LR problems is ALWAYS the smaller value and the reagent responsible for the value is called the LR.
3d. So HCl is the LR and Mg is the excess reagent.
4. You can at this point calculate how much H2 is produced. g H2 = mols H2 x molar mass H2. The problem did not ask for this.
5. The problem asked for how much Mg was used. You know all of the HCl was used so
0.105 mols HCl x (1 mol Mg/2 mols HCl) = 0.105 x 1/2 = approx 0.05 mols Mg used.
Be sure to go back through this and confirm each calculation I've done. I've estimated and rounded off here and there.
mols Mg =
(A) Well, let's use our math skills to figure this one out. First, we need to convert the volume of the hydrochloric acid from centimeters cubed to decimeters cubed. Since 1 decimeter cubed is equal to 1000 centimeters cubed, we can divide 400 centimeters cubed by 1000 to get 0.4 decimeters cubed.
Next, we need to find the number of moles of hydrochloric acid. We can do this by multiplying the volume in decimeters cubed by the concentration in moles per decimeter cubed. So, 0.4 decimeters cubed multiplied by 0.5 moles per decimeter cubed gives us 0.2 moles of hydrochloric acid.
Now, for the magnesium oxide. We have 5 grams of it, and we can use its molar mass to convert grams to moles. The molar mass of magnesium oxide is approximately 40.3 grams per mole. Therefore, 5 grams divided by 40.3 grams per mole gives us approximately 0.124 moles of magnesium oxide.
To determine the limiting reactant, we compare the number of moles of each reactant. We have 0.2 moles of hydrochloric acid and 0.124 moles of magnesium oxide. Since the mole ratio between hydrochloric acid and magnesium oxide is 2:1, the stoichiometry tells us that 0.2 moles of hydrochloric acid would need 0.1 moles of magnesium oxide, but we only have 0.124 moles. Therefore, magnesium oxide is in excess, and hydrochloric acid is the limiting reactant.
(B) Since hydrochloric acid is the limiting reactant, we can use its mole ratio with magnesium oxide to determine how much magnesium actually took part in the reaction. The mole ratio tells us that for every 2 moles of hydrochloric acid, we need 1 mole of magnesium oxide.
Since we have 0.2 moles of hydrochloric acid, we can divide that by 2 to get 0.1 moles of magnesium oxide required. However, remember that we only have 0.124 moles of magnesium oxide. Therefore, 0.124 moles of magnesium actually took part in the reaction.
Now, to convert moles to grams, we multiply the number of moles by the molar mass of magnesium, which is approximately 24.3 grams per mole. Therefore, 0.124 moles of magnesium multiplied by 24.3 grams per mole gives us approximately 3.00 grams of magnesium that actually took part in the reaction.
So, the mass of magnesium that actually took part in the reaction is approximately 3.00 grams.
To determine the excess and limiting reactants in the reaction, we need to calculate the number of moles of each reactant and compare them.
First, let's calculate the number of moles of hydrochloric acid (HCl) present:
Volume of HCl = 400 cm^3 = 400/1000 dm^3 = 0.4 dm^3
Concentration of HCl = 0.5 mol/dm^3
Number of moles of HCl = Concentration x Volume
= 0.5 mol/dm^3 x 0.4 dm^3
= 0.2 mol
Now, let's calculate the number of moles of magnesium oxide (MgO) present:
Molar mass of MgO = 24.31 g/mol + 16.00 g/mol = 40.31 g/mol
Mass of MgO = 5 g
Number of moles of MgO = Mass / Molar mass
= 5 g / 40.31 g/mol
= 0.124 mol
Comparing the number of moles of HCl and MgO, we can see that HCl is in excess since it has a greater number of moles (0.2 mol) compared to MgO (0.124 mol).
To find the mass of magnesium actually took part in the reaction, we need to identify the limiting reactant, which is MgO in this case. The stoichiometry of the balanced equation will help us determine the number of moles of Mg that reacted with the limiting reactant.
The balanced chemical equation for the reaction between MgO and HCl is:
MgO + 2HCl → MgCl2 + H2O
From the equation, we can see that 1 mole of MgO reacts with 2 moles of HCl. Therefore, the number of moles of Mg reacting with the limiting reactant (MgO) is half that of the moles of the limiting reactant.
Number of moles of Mg = 0.124 mol / 2
= 0.062 mol
Finally, let's calculate the mass of magnesium that actually took part in the reaction:
Mass of Mg = Number of moles x Molar mass
= 0.062 mol x 24.31 g/mol
= 1.51 g
Therefore, the mass of magnesium that actually took part in the reaction is 1.51 grams.
To answer these questions, we need to determine the limiting reactant and the excess reactant.
(A) To calculate the limiting reactant, we need to compare the amount of each reactant to the stoichiometric ratio given by the balanced chemical equation. The balanced equation for the reaction between hydrochloric acid (HCl) and magnesium oxide (MgO) is:
HCl + MgO -> MgCl2 + H2O
First, we need to convert the volume of hydrochloric acid from cm³ to dm³:
400 cm³ = 400/1000 dm³ = 0.4 dm³
Next, we convert the concentration of hydrochloric acid from mol/dm³ to mol:
0.5 mol/dm³ x 0.4 dm³ = 0.2 moles of HCl
Now, let's calculate the number of moles of magnesium oxide:
To do this, we need to know the molar mass of MgO, which is 24.31 + 16.00 = 40.31 g/mol.
Given:
Mass of MgO = 5 grams
Number of moles of MgO = Mass of MgO / Molar mass of MgO
= 5 g / 40.31 g/mol
= 0.124 moles of MgO
Now, let's compare the moles of each reactant to the stoichiometry of the balanced equation:
1 mole of HCl reacts with 1 mole of MgO
Since the ratio is 1:1, the moles of HCl and MgO are equal. Therefore, both reactants are in stoichiometric proportions, and there is no limiting or excess reactant in this scenario.
(B) The mass of magnesium that actually took part in the reaction can be determined from the moles of magnesium oxide reacted. We already calculated the moles of MgO to be 0.124 moles.
To determine the mass of magnesium, we need to know the molar mass of Mg, which is 24.31 g/mol.
Mass of magnesium = Number of moles of MgO x Molar mass of Mg
= 0.124 moles x 24.31 g/mol
≈ 3.01 grams
Approximately 3.01 grams of magnesium actually took part in the reaction.