Butane, C4H10 burns in oxygen. How many grams of water vapor, H2O, are produced by the combustion of 580 grams of butane at standard conditions 2C4H10 + 13O2 —-> 8CO2 + 10H2O

The answer to your question is 900 g of water vapor

Explanation:

Data

mass of H₂O = ?

mass of butane = 580 g

Balanced chemical reaction

             2C₄H₁₀ + 13O₂  ⇒  8CO₂  +  10H₂O

Process

1.- Calculate the molar weight of butane and water

Butane (C₄H₁₀) = 2[(12 x 4) + (1 x 10)]

                         = 2[48 + 10]

                         = 2[58]

                         = 116 g

Water (H₂O) = 10[(1 x 2) + (1 x 16)]

                    = 10[2 + 16]

                    = 10[18]

                    = 180 g

2.- Use proportions and cross multiplication to find the mass of water vapor

             116 g of butane 180 g of water

             580 g of butane    x

                x = (580 x 180) / 116

                x = 900 g of water vapor

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.

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.