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lukranit [14]
3 years ago
6

HQ5.49 Use standard enthalpies of formation to calculate the standard enthalpy of reaction for the following reaction: 2 H 2 ​ S

(g) + 3 O 2 ​ (g) → 2 H 2 ​ O(l) + 2 SO 2 ​ (g) ΔH ∘ rxn ​
Chemistry
1 answer:
weqwewe [10]3 years ago
5 0
Answer is: <span>the standard enthalpy of reaction is</span> -1124,2 kJ/mol.
ΔHf(H₂S) = -20.5 kJ/mol
ΔHf(O₂)=0 kJ/mol.
ΔHf(H₂O) = -285.8 kJ/mol.
ΔHf(SO₂) = -296.8 kJ/mol.
ΔHrxn = ∑ΔHf (products of reaction) - ∑ΔHf (reactants).<span>
ΔHrxn - enthalpy change of chemical reaction.
<span>ΔHf - enthalpy of formation of reactants or products.
</span></span>ΔHrxn = (2 · (-285,8) + 2 · (-296,8)) - (2 · (-20,5)) · kJ/mol.
ΔHrxn = -1124,2 kJ/mol.

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In a 0.20 M solution, a weak acid is 3.0% dissociated.(a) Calculate the [H₃O⁺] , pH , [OH⁻], and pOH of the solution.
marin [14]

H₃O⁺ = 6.0×10^{-3}

OH = 1.7×10^{-12}

pH = 2.22

pOH = 11.78

<h3>What is pH?</h3>

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8 0
2 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac
Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

4 0
3 years ago
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Mazyrski [523]

Explanation:

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Prediction, in which the anticipatory deductions based on hypotheses are put forward in testable ways. Statistics can help only a little at this stage of analysis, for predictive insights are often intuitive and creative rather than numerical.
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