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# Consider these reactions, where M represents a generic metal. 2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)ΔH1=−776.0 kJ?

Consider these reactions, where M represents a generic metal.

2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)ΔH1=−776.0 kJ

HCl(g)⟶HCl(aq) ΔH2=−74.8 kJ

H2(g)+Cl2(g)⟶2HCl(g) ΔH3=−1845.0 kJ

MCl3(s)⟶MCl3(aq) ΔH4=−368.0 kJ

Use the given information to determine the enthalpy of the reaction

2M(s)+3Cl2(g)⟶2MCl3(s)

### 1 Answer

- Roger the MoleLv 71 month ago
You need "2 M(s)" on the left. The only given equation that mentions M(s) is the first one, so copy the first given equation:

2 M(s) + 6 HCl(aq) → 2 MCl3(aq) + 3 H2(g) ΔH = −776.0 kJ

You also need "3 Cl2(g)" on the left. The only given equation that mentions Cl2(g) is the third one, so multiply the third given equation by 3:

3 H2(g) + 3 Cl2(g) → 6 HCl(g) ΔH = − 5535.0 kJ

Add the two equations here:

2 M(s) + 6 HCl(aq) + 3 H2(g) + 3 Cl2(g) →

2 MCl3(aq) + 3 H2(g) + 6 HCl(g) ΔH = −776.0 kJ − 5535.0 kJ

Cancel like amounts on opposite sides of the arrow, and do the arithmetic for ΔH:

2 M(s) + 3 Cl2(g) → 2 MCl3(aq) ΔH = − 6311 kJ

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