If 7.50 g of NaClO3 decomposes in the presence of heat to produce 4.12 g
of NaCl, what mass of O2 gas will be produced?
2 NaClO3(s) +2 NaCl(s) + 3 O2(g)
2
UO

3.38 grams of O2 will be produced

Explanation:

Step 1: Data given

Mass of NaClO3 = 7.50 grams

Mass of NaCl = 4.12 grams

Molar mass of NaClO3 = 106.44 g/mol

Molar mass of NaCl = 58.44 g/mol

Step 2: The balanced equation

2NaClO3(s) → 2NaCl(s) + 3O2(g)

Step 3: Calculate moles

Moles NaClO3 = mass NaClO3 / molar mass NaClO3

Moles NaClO3 = 7.50 grams / 106.44 g/mol

Moles NaClO3 = 0.0705 moles

Moles NaCl = 4.12 grams / 58.44 g/mol

Moles NaCl = 0.0705 moles

For 2 moles NaClO3 we'll have 2 moles NaCl and 3 moles O2

For 0.0705 moles NaClO3 we'll have 0.0705 moles NaCl and 3/2 * 0.0705 = 0.10575 moles O2

Step 4: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.10575 moles * 32.0 g/mol

Mass O2 = 3.38 grams O2

3.38 grams of O2 will be produced

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.

c, liquids evaporate, creating a gas