How many grams of nh3 can be produced from the reaction of 28 g of n2 and 25 g of h2

Explanation:

1) Balanced chemical equation

2) Stoichiometric (theoretical ) mole ratios

3) Number of moles of each reactant

4) Limiting reactant

Since the stoichiometry states that 1 mol of N₂ reacts with 3 moles of H₂, the given mass of N₂ will react completely with the given amount of H₂, and the calculations must be done with the 28 g (1 mol) of N₂ as the limiting reactant.

5) Yield

Set the proportion with the mole ratios:

1 mol H₂ / 2 mol NH₃ = 1  mol H₂ / x ⇒ x = 2 mol NH₃

6) Convert to grams

the reaction of 28 g of N₂ with 25 g of H₂ will produce 34 g of NH₃

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.