Suppose 3.50g each of sulfamic acid and sodium nitrite were allowed to react together. What would be the theoretical yield, in moles, of N2.
Equation:
NaNO2 + NH3SO3 -> N2 + H2O + NaHSO4
Notes: The theoretical yield of N2 can be calculated from the amounts of starting materials.
I have no idea how to solve this. Please show work, I want to figure it out and learn, not just know the answer.
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
1. Write and balance the equation. You have that.
2a. Convert grams NaNO2 to mols. mols = g/molar mass = ?
2b. Do the same for sulfamic acid = ?
3a. Using the coefficients in the balanced equation, convert mols NaNO2 to mols N2.
3b. Do the same for sulfamic acid to N2
3c. It is likely that mols N2 will not agree from the 3a and 3b calculations; the correct value in LR problems is ALWAYS the smaller number.
4. The problem doesn't ask for it but if you wanted grams N2, then
grams N2 = mols N2 x molar mass N2.
Limiting reagent problems can be worked with these 4 steps.
Thank you so much DrBob22! Your instructions were very well laid out and extremely helpful. I had initially solved it this way, but I found it odd considering my homework had nothing to do with limiting reactants. Thanks again!
To calculate the theoretical yield of N2 in moles, we need to use the stoichiometry of the balanced equation and the amounts of the reactants given.
1. Determine the molar masses of sulfamic acid (NH3SO3) and sodium nitrite (NaNO2):
- Molar mass of NH3SO3 = 3(1.01 g/mol H) + 1(14.01 g/mol N) + 1(32.06 g/mol S) + 3(16.00 g/mol O) = 97.09 g/mol
- Molar mass of NaNO2 = 1(22.99 g/mol Na) + 1(14.01 g/mol N) + 2(16.00 g/mol O) = 69.00 g/mol
2. Convert the given masses of sulfamic acid and sodium nitrite to moles:
- Moles of NH3SO3 = 3.50 g NH3SO3 / 97.09 g/mol = 0.036 moles
- Moles of NaNO2 = 3.50 g NaNO2 / 69.00 g/mol = 0.051 moles
3. Determine the stoichiometric ratio between sodium nitrite and nitrogen gas (N2) based on the balanced equation:
- From the equation, we see that 1 mole of NaNO2 reacts to produce 1 mole of N2.
4. Calculate the theoretical yield of N2:
- The theoretical yield of N2 in moles is equal to the moles of NaNO2 present, which is 0.051 moles.
Therefore, the theoretical yield of N2 in moles, when 3.50g each of sulfamic acid and sodium nitrite are allowed to react together, is approximately 0.051 moles.