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?
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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?
#1 makes no sense. You have the heat of formation from the constituent elements listed in your question and the question marks by a reaction that is not the formation of water from its constituent elements.
#2.
Write the equation.
Convert 2 g Zn to moles.
Convert 2.5 g AgNO3 to mols.
Using the coefficients in the balanced equation, convert each of the moles into moles of any one (I suggest Ag) of the products.
The smaller number of moles of product formed from either of the reactants will be the limiting reagent; therefore, that will be the number of grams Ag produced. Post your work if you get stuck.
2.0g Zn= 0.0305mol 2.5gAgNO3 =0.0147mol
I am stuck after this. I cannot come to the correct answer.
To calculate the heat of formation of water from its constituent elements, we need to find the ΔH (enthalpy change) for the given reaction:
3 H2 (g) + O3 (g) → 3H2O
First, we can check if we can directly use the values given for the other reactions to obtain the ΔH for this reaction. However, we cannot use the given reactions directly to calculate ΔH for the formation of water because they involve different numbers of moles of reactants and products.
To solve this, we can use a combination of the given reactions to cancel out some of the compounds and get to the desired reaction:
2 H2 (g) + O2 (g) → 2H2O (∆H = -483.6)
3 O2 (g) → 2O3 (g) (∆H = 284.5)
Since we need 3 moles of water in the final reaction, we can multiply the first reaction by 3:
6 H2 (g) + 3 O2 (g) → 6H2O (∆H = -1447.8)
But we require only one mole of O2 in the final reaction, so we need to divide the second reaction by 3:
O2 (g) → (2/3)O3 (g) (∆H = 284.5/3)
Now, we can add these two reactions together:
6 H2 (g) + 3 O2 (g) + O2 (g) → 6H2O + (2/3)O3 (g)
Simplifying the equation:
6 H2 (g) + 4 O2 (g) → 6H2O + (2/3)O3 (g)
Since we want to get rid of O3, we multiply the equation by 3:
6 H2 (g) + 4 O2 (g) → 6H2O + 2O3 (g)
Now, we can cancel O3 on both sides:
6 H2 (g) + 4 O2 (g) → 6H2O
The ΔH for this new equation is the same as the ΔH for the original equation we want to solve.
Since this new equation is a combination of the given reactions, we can calculate the ΔH for this reaction:
ΔH = (-483.6) + (284.5/3)
ΔH = -161.9 kJ (rounded to one decimal place)
Therefore, the heat of formation of water from its constituent elements is -161.9 kJ.
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To find the grams of silver produced in the given reaction, we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed in the reaction and determines the amount of product formed.
We're given 2.0 grams of zinc (Zn) and 2.5 grams of silver nitrate (AgNO3). We can use stoichiometry to find the limiting reagent by calculating the number of moles of each reactant.
The molar mass of Zn is 65.38 g/mol, and the molar mass of AgNO3 is 169.87 g/mol.
Number of moles of Zn = mass of Zn / molar mass of Zn
= 2.0 g / 65.38 g/mol
≈ 0.0305 mol
Number of moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
= 2.5 g / 169.87 g/mol
≈ 0.0147 mol
From the balanced equation, we see that 1 mole of Zn reacts with 2 moles of AgNO3 to produce 2 moles of Ag:
Zn (s) + 2AgNO3 (aq) → 2Ag (s) + Zn(NO3)2 (aq)
Therefore, the mole ratio of Zn to Ag is 1:2.
The number of moles of Ag that can be produced is limited by the mole ratio, so the limiting reagent is AgNO3. This means that all of the AgNO3 will be consumed, and Zn will be in excess.
To calculate the grams of Ag produced, we can use the mole ratio:
Number of moles of Ag = 2 × number of moles of AgNO3
= 2 × 0.0147 mol
= 0.0294 mol
Mass of Ag = molar mass of Ag × number of moles of Ag
= 107.87 g/mol × 0.0294 mol
≈ 3.17 g (rounded to two decimal places)
Therefore, approximately 3.17 grams of silver are produced.