For the reaction 3NO2 + H2O � 2HNO3 + NO, how many grams of HNO3 can form when 1.00 g of NO2 and 2.25 g of H2O are allowed to react?
This is a limiting reagent problem.
1. Write the equation. You have done that and it is balanced.
2. Convert 1.00 g NO2 and 2.25 g H2O to mols.
a. mols NO2 = grams/molar mass = ??
b. mols H2O = grams/molar mass = ??
3. Using the coefficients in the balanced equation, convert mols of what you have (in this case mols NO2 or mols H2O) to mols of what you want (in this case mols HNO3).
3a. using mols NO2 from 2a.
mols HNO3 = mols NO2 x (2 mol HNO3/3 mols NO2) = ??
3b. using mols H2O from 2b.
mols HNO3 = mols H2O x (1 mol H2O/2 mols HNO3) = ??
3c. The smaller of the two answers for HNO3 will be the correct one to use and that is the limiting reagent. The other reagent will be the one that has some material unreacted.
4. Now convert mols of HNO3 to grams HNO3 by
grams HNO3= mols HNO3 x molar mass HNO3.
Post your work if you get stuck.
The above solution assumes the reaction goes to completion from left to right.
1.638 grams
To determine the limiting reagent and calculate the grams of HNO3 formed, follow these steps:
1. Find the molar masses of NO2 and H2O:
- Molar mass of NO2 = 46.01 g/mol
- Molar mass of H2O = 18.02 g/mol
2. Convert grams of NO2 and H2O to moles:
- Moles of NO2 = 1.00 g NO2 / 46.01 g/mol = 0.0217 mol NO2
- Moles of H2O = 2.25 g H2O / 18.02 g/mol = 0.125 mol H2O
3. Determine the limiting reagent:
- From the balanced equation, the stoichiometric ratio of NO2 to HNO3 is 3:2.
- Thus, for every 3 moles of NO2, 2 moles of HNO3 are formed.
- The moles of HNO3 that can be formed from the available NO2 is:
Moles of HNO3 from NO2 = Moles of NO2 x (2 mol HNO3 / 3 mol NO2)
= 0.0217 mol NO2 x (2/3) = 0.0145 mol HNO3
- The moles of HNO3 that can be formed from the available H2O is:
Moles of HNO3 from H2O = Moles of H2O x (1 mol H2O / 2 mol HNO3)
= 0.125 mol H2O x (1/2) = 0.0625 mol HNO3
- Since 0.0145 mol of HNO3 can be obtained from NO2 and 0.0625 mol can be obtained from H2O, the limiting reagent is NO2.
4. Calculate the grams of HNO3 formed from the limiting reagent:
- Grams of HNO3 formed = Moles of HNO3 from the limiting reagent x Molar mass of HNO3
= 0.0145 mol HNO3 x 63.01 g/mol = 0.913 g HNO3 (rounded to 3 decimal places)
Therefore, when 1.00 g of NO2 and 2.25 g of H2O are allowed to react, approximately 0.913 grams of HNO3 can be formed.
To calculate the grams of HNO3 formed, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
1. Write the balanced equation:
3NO2 + H2O → 2HNO3 + NO
2. Calculate the moles of NO2 and H2O:
- Moles of NO2 = 1.00 g NO2 / molar mass of NO2
- Moles of H2O = 2.25 g H2O / molar mass of H2O
3. Use the coefficients in the balanced equation to convert moles of NO2 or H2O to moles of HNO3:
a. Moles of HNO3 (using NO2):
Moles of HNO3 = Moles of NO2 x (2 mol HNO3 / 3 mol NO2)
b. Moles of HNO3 (using H2O):
Moles of HNO3 = Moles of H2O x (1 mol H2O / 2 mol HNO3)
4. Compare the calculated moles of HNO3 from step 3a and 3b. The smaller value represents the limiting reagent.
5. Convert the moles of HNO3 to grams:
Grams of HNO3 = Moles of HNO3 x molar mass of HNO3
Remember to use accurate molar masses for each substance in the calculations.