This article will delve into the stoichiometric calculation of the grams of ammonia (NH3) that completely react with 8.84 grams of nitrogen monoxide (NO). The balanced chemical equation for the reaction is:
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4 NH3(g) + 6 NO(g) → 5 N2(g) + 6 H2O(g)
This equation tells us that 4 moles of ammonia react with 6 moles of nitrogen monoxide to produce 5 moles of nitrogen gas and 6 moles of water. We can use this information to determine the mass of ammonia required to react with a given mass of nitrogen monoxide.
Steps to Calculate the Mass of Ammonia
The following steps outline the calculation:
1. **Convert grams of NO to moles of NO:** This step involves using the molar mass of NO, which is 30.01 g/mol.
2. **Use the mole ratio from the balanced equation to find moles of NH3:** The balanced equation shows that 4 moles of NH3 react with 6 moles of NO. This ratio can be used to convert moles of NO to moles of NH3.
3. **Convert moles of NH3 to grams of NH3:** Using the molar mass of NH3 (17.03 g/mol), we can convert the moles of NH3 to grams.
Detailed Calculation
Let's perform the calculations step by step:
1. **Moles of NO:**
Moles of NO = (Mass of NO) / (Molar mass of NO)
Moles of NO = 8.84 g / 30.01 g/mol
Moles of NO = 0.295 mol
2. **Moles of NH3:**
Moles of NH3 = (Moles of NO) * (Molar ratio of NH3 to NO)
Moles of NH3 = 0.295 mol * (4 mol NH3 / 6 mol NO)
Moles of NH3 = 0.197 mol
3. **Grams of NH3:**
Grams of NH3 = (Moles of NH3) * (Molar mass of NH3)
Grams of NH3 = 0.197 mol * 17.03 g/mol
Grams of NH3 = 3.35 g
Conclusion
Therefore, 3.35 grams of ammonia (NH3) react completely with 8.84 grams of nitrogen monoxide (NO). This calculation demonstrates the application of stoichiometry in chemistry, where the balanced chemical equation provides the mole ratios required for precise mass calculations.