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
not enough nitrogen
a mole of nitrogen is needed for every 3 moles of hydrogen
To determine the limiting reactant, we need to compare the amounts of each reactant to their stoichiometric ratios in the balanced equation.
1. Convert the mass of each reactant to moles:
- N2: 5.42 g / 28.02 g/mol = 0.194 mol
- H2: 5.42 g / 2.02 g/mol = 2.68 mol
2. Calculate the mole ratio between N2 and H2 using the balanced equation:
- According to the balanced equation, the ratio of N2 to H2 is 1:3.
3. Adjust the moles of each reactant based on the stoichiometric ratio:
- N2: 0.194 mol × (1/1) = 0.194 mol
- H2: 2.68 mol × (1/3) = 0.893 mol
Based on the calculations, we can see that the number of moles of nitrogen gas (N2) is less than the number of moles of hydrogen gas (H2). Therefore, the limiting reactant is N2. This means that N2 will be completely consumed in the reaction, and there will be excess H2 remaining after the reaction is complete.