Final answer:
The reaction of 6.30g of ammonia with 1.80g of oxygen at STP would produce 8.29 liters of nitrogen monoxide, determined by using stoichiometry and the molar volume of a gas at STP.
Step-by-step explanation:
The question asks about how many liters of nitrogen monoxide (NO) would be produced from the reaction of 6.30g of ammonia (NH3) with 1.80g of oxygen (O2) at standard temperature and pressure (STP).
To answer this question, we first need the balanced chemical equation. One possible reaction between ammonia and oxygen to produce nitrogen monoxide and water is:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Let us calculate the moles of ammonia (NH3) and oxygen (O2) we start with:
Molar mass of NH3 = 17.03 g/mol
Moles of NH3 = 6.30g / 17.03 g/mol = 0.370 mol
Molar mass of O2 = 32.00 g/mol
Moles of O2 = 1.80g / 32.00 g/mol = 0.056 mol
Using the stoichiometry of the reaction, 4 moles of NH3 react with 5 moles of O2, making NH3 the limiting reagent:
0.370 mol NH3 × (4 mol NO / 4 mol NH3) = 0.370 mol NO
At STP, 1 mole of gas occupies 22.4 liters, so the volume of NO produced is:
Volume of NO = 0.370 mol × 22.4 L/mol = 8.29 L
Therefore, 8.29 liters of nitrogen monoxide would be formed from the reaction at STP.