How many liters of nitrogen oxide at STP are produced from the reaction of 59.g of NH3?
Balanced Equation:
Ammonia reacts with oxygen to produce nitrogen oxide and water.
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
mols NH3 = 59/molar mass NH3 = ?
Convert mols NH3 to mols NO using the coefficients in the balanced equation. That's ?mols NH3 x (4 mols NO/4 mols NH3) = ? x 1/1 = ?mols NO.
Then ? mols NO x 22.4L/mol = x L NO @ STP
To find the number of liters of nitrogen oxide produced, we need to use stoichiometry and convert the given mass of ammonia (NH3) to moles, then use the balanced equation to convert moles of NH3 to moles of nitrogen oxide (NO), and finally convert moles of NO to liters at standard temperature and pressure (STP).
1. Calculate the moles of NH3:
Number of moles = Given mass / Molar mass
The molar mass of NH3 (ammonia) is:
1(N) + 3(H) = 14.01 g/mol + (3 * 1.01 g/mol) = 17.03 g/mol
Moles of NH3 = 59 g / 17.03 g/mol = 3.46 mol
2. Use the balanced equation to determine the moles of NO produced:
According to the balanced equation, 4 moles of NH3 react to produce 4 moles of NO.
So, 3.46 moles of NH3 will produce (3.46 moles * 4 moles of NO) / 4 moles of NH3 = 3.46 moles of NO.
3. Convert moles of NO to the volume at STP:
At STP, 1 mole of any gas occupies 22.4 liters of volume.
Thus, the volume of NO at STP = 3.46 moles * 22.4 L/mol = 77.5 L
Therefore, 59.0 g of NH3 will produce 77.5 L of nitrogen oxide at STP.