If 6.52l of h2 react with excess nitrogen what volume of nh3 is produced?

The volume of NH₃ produced is 3.6 L

Explanation:

Let's make the reaction:

N₂ (g)  +  3H₂ (g)  ⇄  2NH₃ (g)

We have the volume of H₂ which has reacted, but we don't have the mass (neither the moles). We must apply density of H₂ to know this data:

δH₂ (g) =  0.071 g/l

density = mass / volume

0.071 g/L = mass / 6.52L

0.071 g/L . 6.52L = mass → 0.463 g

Molar mass of H₂ = 2 g/m

Mass / Molar mass = moles

0.463 g /2g/m = 0.231 moles

In the reaction ratio between hydrogen and ammonia is 3:2

3 moles of H₂ 2 moles of ammonia

0.231 moles H₂   (0.231 .2)/3 = 0.154 moles

These are the moles of ammonia which are produced, so we have to convert this to mass, to apply NH₃ 's density.

moles . molar mass = mass

Molar mass NH₃ = 17 g/m

0.154 m . 17g/m = 2.62 g

NH₃ density = 0,00073 g/ml

NH₃ density = NH₃ mass / NH₃ volume

NH₃ volume =  NH₃ mass / NH₃ density

NH₃ volume = 2.62 g /0,00073 g/ml

NH₃ volume = 3593.4 mL

3593.4 mL /1000 ≅ 3.6L

i'm pretty sure it is letter c.) car

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.