Ethanol, C2H5OH, is considered a clean fuel because it burns in oxygen to produce carbon dioxide and water with few trace pollutants. If 95.0 g of H2O are produced during the combustion of ethanol, how many grams of ethanol were present at the beginning of the reaction?

80.9 g of ethanol were present at the begining of the reaction

Explanation:

In a combustion reaction you should know that the formed products are water and carbon dioxide. One of the reactants is always oxygen.

The combustion reaction for the ethanol has this equation:

C₂H₅OH (l) + 3O₂(g)  →  2CO₂(g)  +  3H₂O(g)

We determine the moles of produced water

95 g / 18 g/mol = 5.28 moles . Now we can propose this rule of three:

3 moles of water are produced by 1 mol of ethanol

Therefore 5.28 moles of water were produced by (5.28 . 1) /3 = 1.76 moles of ethanol

Finally we should convert the moles to mass in order to find out the grams that were present at the beginning of the reaction

1.76 mol . 46 g /1 mol = 80.9 g

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.