N2 + 3 h2 → 2 nh3 . what is the percent yield if 45.5 grams of nh3 is produced upon reaction of 13.0 moles n2 and 29.0 moles h2?

13.82 %.

Explanation:

% yield = [(actual mass of NH₃) / (theoretical mass of NH₃)] x 100.

The actual mass of NH₃ = 45.5 g.

Using cross multiplication:

3.0 moles of H₂ produces → 2.0 moles of NH₃, from the stichiometry,

29.0 moles of H₂ produces → ??? moles of NH₃.

∴ The number of moles of NH₃ produced upon reaction of (13.0 moles N₂ and 29.0 moles H₂) = (2.0)(29.0) / (3.0) = 19.333 mol.

∴ % yield = [(actual mass of NH₃) / (theoretical mass of NH₃)] x 100 = [(45.5 g) / (329.266 g)] x 100 = 13.82 %.

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

wanswer:

explanation: