Calculate the mass of butane needed to produce 59.8 g of carbon dioxide. express your answer to three digits and include the appropriate units.

We need 19.8 grams of butane

Explanation:

Step 1: Data given

Mass of CO2 produced = 59.8 grams

Molar mass CO2 = 44.01 g/mol

Molar mass butane =     58.12 g/mol

Step 2: The balanced equation

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Step 3: Caclulate moles CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 59.8 grams / 44.01 g/mol

Moles CO2 = 1.36 moles

Step 4: Calculate moles butane

For 2 moles butane we need 13 moles O2 to produce 8 moles CO2 and 10 moles H2O

For 1.36 moles CO2 we need 1.36/4 = 0.34 moles butane

Step 5: Calculate mass butane

Mass butane = moles butane * molar mass butane

Mass butane = 0.34 moles * 58.12 g/mol

Mass butane = 19.76 grams ≈ 19.8 grams

We need 19.8 grams of butane

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.