Calculate the heat of formation of water from its constituent elements from the following information:
3 H2 (g) + O3 (g) → 3H2O ∆H = ?
2 H2 (g) + O2 (g) → 2H2O ∆H = - 483.6
3 O2 (g) → 2O3 (g) ∆H = 284.5
You are given 2.0 grams of zinc and 2.5 grams of silver nitrate and are told they react in the following manner: Zn¬(s) + 2AgNO3(aq) → 2Ag(s) + Zn(NO3)2 (aq). How many grams of silver are produced?
answered below.
To calculate the heat of formation of water from its constituent elements, we need to use the given enthalpy change (∆H) values for the given reactions. The key is to manipulate and combine these reactions to cancel out the common elements (such as H2 and O2) until we are left with the desired reaction.
Let's start by looking at the given reactions:
1. 3 H2 (g) + O3 (g) → 3H2O ∆H = ?
2. 2 H2 (g) + O2 (g) → 2H2O ∆H = -483.6
3. 3 O2 (g) → 2O3 (g) ∆H = 284.5
By observing the given reactions, we can see that the reaction we are looking for (the formation of water from its elements) involves H2 and O2. Our goal is to eliminate the H2 and O2 from the reactions by manipulating them.
Let's start by manipulating Equation 1 to get rid of the O3 term by combining it with Equation 3:
1. 3 H2 (g) + O3 (g) → 3H2O ∆H = ?
3. 3 O2 (g) → 2O3 (g) ∆H = 284.5
By canceling out the O3 term, we get:
1. 3 H2 (g) + 3 O2 (g) → 3H2O ∆H = ?
Now, let's manipulate Equation 2 to eliminate the O2 term by multiplying it by 3 and flipping its sign to match Equation 1:
2. 2 H2 (g) + O2 (g) → 2H2O ∆H = -483.6
Multiplying Equation 2 by 3 and flipping the sign, we get:
2. 6 H2 (g) + 3 O2 (g) → 6H2O ∆H = 1450.8
Now, we can add Equations 1 and 2 together to cancel out the common H2O term:
1. 3 H2 (g) + 3 O2 (g) → 3H2O ∆H = ?
2. 6 H2 (g) + 3 O2 (g) → 6H2O ∆H = 1450.8
+ (addition)
------------------------
3. 9 H2 (g) + 6 O2 (g) → 9H2O ∆H = ?
Now, we have the desired reaction:
3 H2 (g) + 3 O2 (g) → 3H2O ∆H = ?
We'll refer to this equation as Equation 4.
According to the law of conservation of energy, if two reactions are added together to form a third reaction, their enthalpy changes (∆H) can also be added together.
So, to calculate the ∆H for Equation 4, we need to add the ∆H values from Equations 1 and 2:
∆H4 = ∆H1 + ∆H2
However, before we can do that, we need to manipulate Equation 2 by multiplying it by 3 to balance the number of H2 terms:
2. 6 H2 (g) + 3 O2 (g) → 6H2O ∆H = -1450.8
By multiplying Equation 2 by 3, we get:
2. 6 H2 (g) + 3 O2 (g) → 6H2O ∆H = -1450.8 * 3 = -4352.4
Now, we can add the ∆H values:
∆H4 = ∆H1 + ∆H2 = 284.5 + (-4352.4) = -4067.9 kJ/mol
Therefore, the heat of formation of water from its constituent elements is -4067.9 kJ/mol.
Now, let's move on to the second question:
To determine how many grams of silver are produced in the reaction Zn(s) + 2AgNO3(aq) → 2Ag(s) + Zn(NO3)2(aq), we can use stoichiometry and molar ratios.
First, we need to determine the limiting reactant by comparing the number of moles of zinc (Zn) and silver nitrate (AgNO3) used.
Using the given masses, we can calculate the number of moles of each reactant.
Number of moles of Zn = mass of Zn / molar mass of Zn
Number of moles of Zn = 2.0 g / 65.38 g/mol (molar mass of Zn) = 0.0306 mol
Number of moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
Number of moles of AgNO3 = 2.5 g / (107.87 g/mol + (14.01 g/mol + 3 * 16.00 g/mol)) (molar mass of AgNO3) = 0.0121 mol
The stoichiometric ratio between Zn and Ag is 1:2, meaning that for every 1 mole of Zn reacted, 2 moles of Ag are produced. In this case, we have excess AgNO3, so Zn is the limiting reactant.
From the balanced equation, we know that 1 mole of Zn produces 2 moles of Ag.
Therefore, the number of moles of Ag produced is 2 * 0.0306 mol = 0.0612 mol.
Finally, we can calculate the mass of Ag produced using the molar mass of Ag:
Mass of Ag = number of moles of Ag * molar mass of Ag
Mass of Ag = 0.0612 mol * 107.87 g/mol (molar mass of Ag) = 6.61 g (rounded to two decimal places)
Therefore, 6.61 grams of silver are produced in the reaction.