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

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

d.) circle motion requires a force to give it a centripetal acceleration.