how many liters of nitrogen monoxide are formed if 8.75 g of ammonia are reacted in the presence of excess oxygen?
Here is a solved example stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of liters of nitrogen monoxide formed, you need to use stoichiometry and convert the given mass of ammonia to moles and then calculate the moles of nitrogen monoxide produced. Finally, convert the moles of nitrogen monoxide to liters using the ideal gas law.
Here's the step-by-step explanation:
1. Begin by writing out the balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO):
4NH3 + 5O2 -> 4NO + 6H2O
2. Calculate the molar mass of ammonia (NH3) by adding up the atomic masses of nitrogen (N) and three hydrogen (H) atoms:
Molar mass of NH3 = (1 atomic mass of N) + (3 atomic masses of H)
= 14.01 g/mol + (3 × 1.01 g/mol)
= 17.03 g/mol
3. Convert the given mass of ammonia (8.75 g) to moles by using the molar mass of NH3:
Moles of NH3 = Mass of NH3 / Molar mass of NH3
= 8.75 g / 17.03 g/mol
≈ 0.514 mol
4. From the balanced equation in step 1, we see that 4 moles of NH3 are required to produce 4 moles of NO. Therefore, using the ratio from the balanced equation, we can determine the moles of NO produced:
Moles of NO = Moles of NH3
= 0.514 mol
5. Now, using the ideal gas law (PV = nRT), we can convert the moles of NO to liters:
First, determine the molar volume of an ideal gas at standard temperature and pressure (STP). At STP, 1 mole of any ideal gas occupies 22.414 liters.
Moles of NO x Molar volume at STP = Volume of NO
Volume of NO = 0.514 mol x 22.414 L/mol
≈ 11.5 L
Therefore, approximately 11.5 liters of nitrogen monoxide will be formed when 8.75 grams of ammonia are reacted in the presence of excess oxygen.