N2 + 3H2 Right arrow. 2NH3 What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3? Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..

53.4%

Explanation:

N₂ + 3H₂ → 2NH₃

Given that:

26.3 g of H₂ react with N₂ to produce 79.0 g of NH₃

Then:

The number of moles of H₂ = 26.3g of H₂ * (1 mol of H₂/ 2.02g of H₂)

= 13.05 mol of H₂

The number of moles of NH₃ = 13.05 mol of H₂ * ( 2mol of NH₃/ 3 mol of H₂)

= 8.697 mol of NH₃

The mass of NH₃ = 8.697 mol of NH₃ *( 17.04g of NH₃/ 1 mol of NH₃) = 148.1 g of NH₃

The percent yield = actual yield/ theoretical yield * 100%

The percent yield = ( 79.0 g/ 148.1 g )* 100%

The percent yield ≅ 53.4 %

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )