5.42 grams H2=2.71 moles
YOu need 1/3 of that of N2
moles N2=1/3*2.71=.89
grams N2=.89*28= way more than the 5.42 grams you have, so N2 is the limiting reageant.
N2(g) + 3 H2(g) → 2 NH3(g) [balanced]
If 5.42 g of nitrogen gas are reacted with 5.42 g of hydrogen gas, which of the reactants is the limiting reactant? Use
the molar mass data below if necessary.
Molar mass of N2 = 28.02 g/mol
Molar mass of H2 = 2.02 g/mol
Molar mass of NH3 = 17.04 g/mo
2 answers
Convert gram values given to moles. (moles = mass / f.wt), then divide the mole values by respective coefficients... the smaller number is the limiting reagent. Moles N2 = 5.42/28.02 = 0.193; Moles H2 = 5.42/2.02 = 2.68 => compare N2(0.193/1) and H2(2.68/3) = 0.894... N2 is the limiting reagent & H2 will be in excess.