Question

Given:
4NH3 (g) + 5O2 (g) ------> 4NO (g) + 6H2O (g)
I start with:
1.4 g of NH3
2.85 g O2
I eventually make 1.6 g H2O

What is the theoretical yield of H2O?
What is the percent yield?

Answer 1

1. Determine the limiting reagent.
2. Use moles of limiting reagent to find the maximum possible moles of H2O (use the stoichiometric coefficients).
3. Convert moles of H2O to mass

4. then: \(percent~yield=\dfrac{actual~yield}{theorerical~yield}*100\%\)

Answer 2

Since O2 is the limiting reagent, does this mean that I should multiply the .089 moles of O2 by (5/6) to get .074 moles of H2O? If so Is that 1.335 g of H2O thus 83.44% yield?

Answer 3

yep, thats the right process. i didn't check your actual calculations though, so you may want to double-check that if you think it may have mistakes.