3.25g of zinc was added to 50cm^3 of a 4.0 molar dm^-3 HCl. When the reaction was completed the temperature rose by 28 degrees celcius.
(a) Is the reaction exothermic or endothermic?
(b)All the zinc dissolve in this reaction. How many moles of zinc reacted?
(c) How many moles of acid are there in 50cm^3 of a 4.0 molar dm^-3 HCl.
(d) Write a balance equation for the reaction b/w zinc and HCL.
(e) Which of the two is used in excess? Why?
(g) Calculate the heat change when 3.25g of zinc reacts with 50 cm^3 of a 4.0 molar dm^-3 HCl.
(h) calculate the enthalpy change for the reaction between zinz and HCl.
The reaction is exothermic because heat is evolved.
mols Zn = grams/atomic mass = ?
mols acid = M x L = ?
Zn + HCl ==> ZnCl2 + H2 You balance.
The reagent in excess is not the limiting reagent.
q = mass H2O x specific heat H2O x delta T.
h) dH = q
Note that this is in J/rxn and not J/mol or kJ/mol
Also, not the correct spelling of celsius.
(a) To determine if the reaction is exothermic or endothermic, we need to consider the temperature change. In this case, the temperature rose by 28 degrees Celsius, indicating that heat was released during the reaction. Therefore, the reaction is exothermic.
(b) To calculate the number of moles of zinc reacted, we need to use its molar mass. The molar mass of zinc (Zn) is 65.38 g/mol.
First, let's convert the mass of zinc given (3.25 g) to moles:
moles of zinc = mass of zinc / molar mass of zinc
moles of zinc = 3.25 g / 65.38 g/mol
moles of zinc ≈ 0.0497 mol
Therefore, approximately 0.0497 moles of zinc reacted.
(c) To calculate the number of moles of acid in 50 cm^3 of a 4.0 molar dm^-3 HCl, we need to use the volume and molarity provided.
Molarity (M) = moles of solute / volume of solution (in liters)
4.0 mol/dm^3 = moles of HCl / (50 cm^3 ÷ 1000 cm^3/dm^3)
Rearranging the equation to solve for moles of HCl:
moles of HCl = Molarity × Volume (in liters)
moles of HCl = 4.0 mol/dm^3 × (50 cm^3 ÷ 1000 cm^3/dm^3)
moles of HCl = 4.0 × 0.05 mol
moles of HCl = 0.20 mol
Therefore, there are 0.20 moles of acid (HCl) in 50 cm^3 of a 4.0 molar dm^-3 HCl.
(d) The balanced equation for the reaction between zinc (Zn) and hydrochloric acid (HCl) is:
Zn + 2HCl → ZnCl2 + H2
(e) To identify which reactant is in excess, we need to compare the stoichiometry of the balanced equation. The stoichiometry ratio between zinc (Zn) and hydrochloric acid (HCl) is 1:2.
From part (b), we determined that 0.0497 moles of zinc reacted. Since the stoichiometry ratio is 1:2, we need twice as many moles of hydrochloric acid to completely react with the zinc. Therefore, we need 0.0497 × 2 = 0.0994 moles of HCl.
From part (c), we determined that there are 0.20 moles of HCl available. Since the actual amount of HCl exceeds the required amount (0.0994 moles), HCl is in excess.
(g) To calculate the heat change when 3.25g of zinc reacts with 50 cm^3 of a 4.0 molar dm^-3 HCl, we can use the equation:
heat change = specific heat capacity × mass × temperature change
The specific heat capacity of the system is assumed to be constant.
Specific heat capacity is typically given in J/g°C. However, since the mass of zinc (given in grams) and the volume of HCl (given in cm^3) are provided, we should first convert the volume to moles and then calculate the heat change in joules.
First, calculate the volume of HCl using the molarity and conversion factor from cm^3 to dm^3:
volume of HCl = 50 cm^3 ÷ 1000 cm^3/dm^3
volume of HCl = 0.050 dm^3
The moles of HCl reacting with zinc (given in (b)) is approximately 0.0497 mol.
Next, calculate the heat change using the equation mentioned above:
heat change = specific heat capacity × mass of zinc × temperature change
Assuming a specific heat capacity of 4.18 J/g°C, we have:
heat change = 4.18 J/g°C × 3.25 g × 28 °C
heat change ≈ 316.78 J
Therefore, the heat change when 3.25 g of zinc reacts with 50 cm^3 of a 4.0 molar dm^-3 HCl is approximately 316.78 J.
(h) To calculate the enthalpy change (ΔH) for the reaction between zinc and HCl, we can use the heat change calculated in part (g) and the number of moles of zinc calculated in part (b).
Enthalpy change (ΔH) = heat change / moles of zinc reacted
ΔH = 316.78 J / 0.0497 mol
ΔH ≈ 6376.53 J/mol
Therefore, the enthalpy change for the reaction between zinc and HCl is approximately 6376.53 J/mol.
(a) To determine whether the reaction is exothermic or endothermic, we need to analyze the change in temperature. In this case, the temperature rose by 28 degrees Celsius. If the temperature increases, it indicates that heat is being released, making the reaction exothermic.
(b) To find the number of moles of zinc reacted, we can use the formula:
Moles = Mass / Molar mass
The molar mass of zinc is 65.38 g/mol. Therefore,
Moles of zinc = 3.25 g / 65.38 g/mol = 0.0497 mol (rounded)
(c) To calculate the number of moles of acid in 50 cm^3 of a 4.0 molar dm^-3 HCl solution, we need to use the formula:
Moles = Concentration x Volume
The given concentration is 4.0 molar dm^-3. However, the volume is given in cm^3, so we need to convert it to dm^3 by dividing by 1000.
Volume in dm^3 = 50 cm^3 / 1000 = 0.05 dm^3
Now, we can calculate the moles:
Moles of acid = 4.0 mol/dm^3 x 0.05 dm^3 = 0.2 mol
(d) The balanced chemical equation for the reaction between zinc and HCl can be written as:
Zn + 2HCl -> ZnCl2 + H2
(e) To determine which substance is in excess, we need to compare the stoichiometric ratio of zinc and HCl. From the balanced equation, we can see that 1 mol of zinc reacts with 2 mol of HCl.
We already calculated that 0.0497 mol of zinc reacted. Now, let's calculate how many moles of HCl would react completely with that amount of zinc:
Moles of HCl = 0.0497 mol x (2 mol HCl / 1 mol Zn) = 0.0994 mol
However, we previously found that there are 0.2 mol of HCl in 50 cm^3. Since 0.0994 mol of HCl is less than 0.2 mol, we can conclude that HCl is in excess.
The reason HCl is used in excess is to ensure that all the zinc reacts completely, maximizing the efficiency of the reaction and preventing any unreacted zinc from remaining.
(g) To calculate the heat change when 3.25g of zinc reacts with 50 cm^3 of a 4.0 molar dm^-3 HCl, we can use the formula:
Heat change (q) = Moles of limiting reactant x Enthalpy change per mole
Since we have already determined that zinc is the limiting reactant and calculated the moles of zinc (0.0497 mol), we can proceed to calculate the heat change.
The enthalpy change per mole of zinc reacting with HCl is not provided in the question. Therefore, we cannot directly calculate the heat change (q) without additional information.
(h) Similarly to question (g), to calculate the enthalpy change for the reaction between zinc and HCl, we need the molar enthalpy change (ΔH) per mole of zinc reacting with HCl. This information is not provided in the question, so we cannot directly calculate the enthalpy change without additional information.