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The burning of 80.3 g of SiH4 at constant pressure gives off 3790 kJ of heat. Calculate △H for this reaction. SiH4(g) + 2O2(g) ⟶ SiO2(s) + 2H2O(l).
A) -1520 kJ/mol rxn.
B) - 47.2 kJ/mol rxn.
C) - 4340 kJ/mol rxn.
D) -2430 kJ/mol rxn.
E) + 4340 kJ/mol rxn.

Respuesta :

Answer  : The value of ΔH for this reaction is, -1516 kJ/mol

Explanation :

First we have to calculate the moles of [tex]SiH_4[/tex]

[tex]\text{Moles of }SiH_4=\frac{\text{Mass of }SiH_4}{\text{Molar mass of }SiH_4}[/tex]

[tex]\text{Moles of }SiH_4=\frac{80.3g}{32.12g/mol}[/tex]

[tex]\text{Moles of }SiH_4=2.5mol[/tex]

Now we have to calculate the ΔH for this reaction.

As, 2.5 mole of [tex]SiH_4[/tex] react to gives heat = -3790 kJ

So, 1 mole of [tex]SiH_4[/tex] react to gives heat = [tex]-\frac{3790kJ}{2.5mol}[/tex]

                                                                                = -1516 kJ/mol

Therefore, the value of ΔH for this reaction is, -1516 kJ/mol

The value of ΔH for this reaction is -1516 kJ/mol.

Calculation of the value of ΔH:

Before this we have to calculate the moles of SiH4

Moles = Mass / molar mass

= 80.3 / 32.12

= 2.5 mol

Now the value should be

= -3790 / 2.5 mol

= -1516 kJ/mol

hence, we can conclude that The value of ΔH for this reaction is -1516 kJ/mol.

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