1) A 10g sample of H2(g) reacts with a 22g sample of O2(g) according to the equation: _H2(3)+ -_02(3)→ _H,0(1)

Which reactant is limiting? Which is in excess?

H₂ is excess reactant and O₂ the limiting reactant

Explanation:

Based on the chemical reaction:

2H₂(g) + O₂(g) → 2H₂O

2 moles of H₂ react per mole of O₂

To find limiting reactant we need to convert the mass of each reactant to moles:

Moles H₂ -Molar mass: 2.016g/mol-:

10g H₂ * (1mol / 2.016g) = 4.96 moles

Moles O₂ -Molar mass: 32g/mol-:

22g O₂ * (1mol / 32g) = 0.69 moles

For a complete reaction of 0.69 moles of O₂ are needed:

0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂

As there are 4.96 moles,

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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