Respuesta :
We must first balance the equation, we have two atoms of nitrogen and hydrogen in the reactants, so we must put the coefficient 2 on the side of the products to balance the equation. The balanced equation of the reaction will be:
[tex]N_2+H_2\rightarrow2NH[/tex]Now, to find the grams of NH that will be produced, we will follow these steps:
1. We find the moles of N2 and H2 by dividing the grams by the molar mass of each element.
Molar Mass N2=28.0g/mol
Molar Mass H2=2.0g/mol
2. By stoichiometry we find out which is the limiting reactant, we see that for each mole of H2 one mole of N2 reacts, therefore, the limiting reactant will be the one with the least number of moles. The other reactant will be the excess reactant
3. We find the moles of NH using the limiting reactant.
4. We find the grams of NH by multiplying the moles by the molar mass of NH.
5. We find how much of the excess reactant is left by subtracting the initial moles minus the reacting moles.
Let's proceed with the calculations
1. Moles of N2 and H2
[tex]\begin{gathered} molH_2=givengH_2\times\frac{1molH_2}{MolarMassgH_2} \\ molH_2=45.0gH_2\times\frac{1molH_2}{2.0gH_2}=22.5molH_2 \end{gathered}[/tex][tex]\begin{gathered} molN_2=givengN_2\times\frac{1molN_2}{MolarMassgN_2} \\ molN_2=35.0gN_2\times\frac{1molN_2}{28.0gN_2}=1.25molN_2 \end{gathered}[/tex]2. Limiting reactant
We see that there are 22.5 moles of H2 and 1.25 moles of N2. The ratio N2 to H2 is 1/1, therefore, the limiting reactant will be the one with fewer moles, since it will be consumed faster. Therefore, the limiting reactant is nitrogen.
3. Moles of NH
The ratio NH to N2(Limiting reactant) is 2/1, therefore the moles of NH produced will be:
[tex]\begin{gathered} molNH=givenmolN_2\times\frac{2molNH}{1molN_2} \\ molNH=1.25molN_2\times\frac{2molNH}{1molN_{2}}=2.5molNH \end{gathered}[/tex]4. Mass of NH
[tex]\begin{gathered} MassNH=givenmolNH\times\frac{MolarMass,gNH}{1molNH} \\ MassNH=2.5molNH\times\frac{286gNH}{1molNH}=15.0gNH \end{gathered}[/tex]5. Excess of H2
[tex]\begin{gathered} molH_2(LeftOver)=molH_2initial-molH_2react \\ molH_2(LeftOver)=22.5H_2-1.25molhH_2=21.25molH_2 \end{gathered}[/tex]The mass of H2 left over will be:
[tex]\begin{gathered} MassH_2=21.25molH_2\times\frac{MolarMass,gH_2}{1mol} \\ MassH_2=21.25molH_2\times\frac{2.0gH_2}{1molH_2}=42.5gH_2 \end{gathered}[/tex]Answer: 15.0 grams of NH will be produced and the excess will be 42.5g of H2
