(Solved): olem 7.62 - Enhanced - with Feedback Correct The mole-mole ratio is given by the coefficients in th ...

olem 7.62 - Enhanced - with Feedback Correct The mole-mole ratio is given by the coefficients in the balanced equation $$\frac{3\mathrm{mol}{\mathrm{H}}_2}{1\mathrm{mol}{\mathrm{N}}_2}$$ To use this mole-mole ratio to convert from the given mass of $${\mathrm{N}}_2$$ to the mass of $${\mathrm{H}}_2$$ required, $$2.05\mathrm{g}$$ of $${\mathrm{N}}_2$$ must be converted to moles ff used to find the moles of $${\mathrm{H}}_2$$ required from this quantity of $${\mathrm{N}}_2$$. Finally, use the molar mass of $${\mathrm{H}}_2$$ to convert to grams of $${\mathrm{H}}_2$$. In summary, $$\text{ grams of }{\mathrm{H}}_2\text{ produced }=2.05\mathrm{g}{\mathrm{N}}_2\times \frac{1\text{ por }{\mathrm{N}}_7}{28.02\mathrm{g}}$$ Part C How many grams of $${\mathrm{NH}}_3$$ can be produced from $$13.2{\mathrm{g}}_{\text{of }}{\mathrm{H}}_2$$ ? Express your answer with the appropriate units.