Answer:
[tex]\boxed{\text{No}}[/tex]
Explanation:
This is a disguised limiting reactant problem.
We are given the masses of two reactants and asked to determine if we have enough to make a given amount of product.
1. Calculate the mass of NH₃ needed.
[tex]\text{Mass of NH}_{3} = \text{70 batches} \times \dfrac{\text{175 g}}{\text{1 batch}} = \text{12 250 g}[/tex]
2. Assemble the information
We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 28.02 2.016 17.03
N₂ + 3H₂ ⟶ 2NH₃
Mass/g: 22 000 1000 12 250
3. Calculate the moles of NH₃ needed
[tex]\text{Moles of NH$_{3}$} = \text{12 250 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{719.3 mol}[/tex]
4. Calculate the moles of each reactant
[tex]\text{Moles of N$_{2}$} = \text{22 000 g} \times \dfrac{\text{1 mol N$_{2}$}}{\text{28.02 g N$_{2}$}} = \text{785.2 mol NH$_{3}$}\\\text{Moles of H$_{2}$} = \text{1000 g} \times \dfrac{\text{1 mol H$_{2}$}}{\text{2.016 g H$_{2}$}} = \text{496.0 mol H$_{2}$}[/tex]
5. Calculate the moles of NH₃ from each reactant
[tex]\textbf{From N$_{2}$:}\\\text{Moles of NH$_{3}$} = \text{785.2 mol N$_{2}$} \times \dfrac{\text{2 mol NH$_{3}$}}{\text{1 mol N$_{2}$}} = \textbf{1570 mol NH$_{3}$}\\\textbf{From H$_{2}$:}\\\text{Moles of NH$_{3}$} = \text{496.0 mol H$_{2}$} \times \dfrac{\text{2 mol NH$_{3}$}}{\text{3 mol H$_{2}$}} = \textbf{330.7 mol NH$_{3}$}\\\text{They need to produce 719.3 mol of ammonia, but they don't}\\\text{have enough hydrogen,}[/tex][tex]\boxed{\textbf{No,}}\\\text{They will not be able to make enough ammonia to fill the order.}[/tex]
