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How much heat is absorbed in the complete reaction of 3.00 grams of SiO2 with excess carbon in the reaction SiO2(g) + 3C(s) → SiC(s) + 2CO(g) ΔH° = 624.7 kJ/mol?

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maacastrobr

Answer:

31.24 kJ

Explanation:

  • SiO₂(g) + 3C(s) → SiC(s) + 2CO(g)        ΔH° = 624.7 kJ/mol

First we convert 3.00 grams of SiO₂ to moles, using its molar mass:

  • 3.00 g SiO₂ ÷ 60.08 g/mol = 0.05 mol

Now we calculate the heat absorbed, using the given ΔH°:

If the complete reaction of 1 mol of SiO₂ absorbs 624.7 kJ, then with 0.05 mol:

  • 0.05 mol * 624.7 kJ/mol = 31.24 kJ of heat would be absorbed.
pstnonsonjoku

The heat absorbed by 3.00 grams of SiO2 is 31.17 kJ/mol.

The thermochemical equation is;

SiO2(g) + 3C(s) → SiC(s) + 2CO(g) ΔH° = 624.7 kJ/mol

The number of moles of SiO2 = 3.00 g/60.08 g/mol = 0.0499 moles

From the reaction equation;

1 mole of SiO2 absorbs 624.7 kJ/mol

0.0499 moles SiO2 absorbs  0.0499 moles ×  624.7 kJ/mol/1 mole

= 31.17 kJ/mol

Learn more: https://brainly.com/question/2192784

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