An 84-mg sample of a compound is found to contain 36 mg of carbon, 3 mg of hydrogen, 21 mg of nitrogen, and 24 mg of oxygen. If
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Answer:
94,4 kJ of heat transferred.
10,5 g of CH₄ gas are produced.
304 kJ are released.
Explanation:
The reaction is:
2 CH₃OH(g) → 2 CH₄(g) + O₂(g) ΔH = +252 kJ
24,0 g of CH₃OH(g) are:
24,0g CH₃OH× = 0,749 moles of CH₃OH(g)
As two moles of CH₃OH(g) transferred +252 kJ, 0,749 moles of CH₃OH(g) transferred:
0,749 molesCH₃OH× = <em>94,4 kJ</em>
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When 2 moles of CH₄ are produced, the enthalpy change is +252 kJ. 82,6 kJ are:
82,6kJ×= 0,656 moles of CH₄
0,656 moles of CH₄ are:
0,656 moles of CH₄× = <em>10,5 g of CH₄</em>
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38,7 g of CH₄(g) are:
38,7g CH× = 2,41 moles of CH₄(g)
As two moles of CH₄ released 252 kJ, 2,41 moles of CH₄(g) released:
2,41 moles CH₄× = <em>304 kJ</em>
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I hope it helps!