Answer:
ΔH = -20kJ
Explanation:
The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:
H₂(g) + S(g) → H₂S(g)
Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:
<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
The sum of -(1) + (2) + (3) gives:
<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>
<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>
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I hope it helps!
The name is sodium chloride. This is an ionic compound so you use the ionic naming system which is the name of the cation followed by the name of the anion. I hope this helps. Let me know if anything is unclear.
Answer:
<em>The grams of hydrogen gas that are released is 0.343 gram</em>
Explanation:
<em>Given that:</em>
<em>2 Na + 2h₂0 ⇒ 2Na0H +N₂</em>
<em>Molecules of Na = 7.9/2.3 = 0.343 mol Na</em>
<em>Now,</em>
<em>2 molecules of Na produces one mole h₂</em>
<em>0.343 mol of Na produce 0.343/2 = 0.1717 mol h₂</em>
<em>h2 = molecules * mw (molecular weight) =0.1717 * 2 =0.343 grams</em>
<em>Therefore the grams of hydrogen gas released is = 0.343 grams</em>
Answer:
We'll have 8.0 moles Fe3O4 and 4.0 moles CO2
Explanation:
Step 1: data given
Number of moles Fe2O3 = 12.0 moles
Number of moles CO = 12.0 moles
Step 2: The balanced equation
3Fe2O3 +CO → 2Fe3O4 + CO2
Step 3: Calculate the limiting reactant
For 3 moles Fe2O3 we need 1 mol CO to produce 2 moles Fe3O4 and 1 mol CO2
Fe2O3 is the limiting reactant. It will completely be consumed (12.0 moles).
CO is in excess. There will react 12.0 / 3 = 4.0 moles
There will remain 12.0 - 4.0 = 8.0 moles
Step 4: Calculate moles products
For 3 moles Fe2O3 we need 1 mol CO to produce 2 moles Fe3O4 and 1 mol CO2
For 12.0 moles Fe2O3 we'll have 2/3 * 12.0 = 8.0 moles Fe3O4
For 12.0 moles Fe2O3 we'll have 12.0 / 3 = 4.0 moles CO2
We'll have 8.0 moles Fe3O4 and 4.0 moles CO2
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