Calculate the moles of oxygen gas needed t burn 1.22 moles of ammonia, NH3(g). Find the moles of nitrogen dioxide gas and gaseous water produced.
Just follow the steps.
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The balanced chemical equation for the combustion of ammonia is:
4NH3(g) + 5O2(g) → 4NO2(g) + 6H2O(g)
According to the equation, 4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO2 and 6 moles of H2O.
Given that you have 1.22 moles of NH3, we can use the stoichiometric ratio to determine the amount of O2 needed.
1.22 moles NH3 × (5 moles O2 / 4 moles NH3) = 1.525 moles O2
Therefore, 1.525 moles of oxygen gas are needed to burn 1.22 moles of ammonia.
Using the same stoichiometric ratio, we can calculate the moles of nitrogen dioxide (NO2) and water (H2O) produced.
1.525 moles O2 × (4 moles NO2 / 5 moles O2) = 1.22 moles NO2
1.525 moles O2 × (6 moles H2O / 5 moles O2) = 1.83 moles H2O
Therefore, 1.22 moles of nitrogen dioxide gas and 1.83 moles of gaseous water are produced when 1.22 moles of ammonia are burned.
To determine the moles of oxygen gas needed to burn ammonia (NH3(g)), we need to balance the chemical equation for the reaction.
The balanced equation for the combustion of ammonia is as follows:
4NH3(g) + 5O2(g) → 4NO2(g) + 6H2O(g)
From the equation, we can see that it takes 5 moles of O2 to react with 4 moles of NH3.
Given that we have 1.22 moles of NH3, we can set up a ratio to determine the amount of O2 needed.
1.22 moles NH3 / 4 moles NH3 = x moles O2 / 5 moles O2
Cross-multiplying and solving for x, we have:
x = (1.22 moles NH3 * 5 moles O2) / 4 moles NH3 = 1.525 moles O2
Therefore, we need 1.525 moles of O2 to burn 1.22 moles of NH3.
Now, to find the moles of nitrogen dioxide (NO2) and gaseous water (H2O) produced, we can use the coefficients from the balanced equation.
From the balanced equation, we see that for every 4 moles of NH3 that react, we produce 4 moles of NO2 and 6 moles of H2O.
Using the same ratio approach as before, we can calculate:
1.22 moles NH3 / 4 moles NH3 = x moles NO2 / 4 moles NO2
Cross-multiplying and solving for x, we have:
x = (1.22 moles NH3 * 4 moles NO2) / 4 moles NH3 = 1.22 moles NO2
Therefore, 1.22 moles of NO2 are produced when 1.22 moles of NH3 are burned.
Similarly, we can calculate the moles of gaseous water produced:
1.22 moles NH3 / 4 moles NH3 = x moles H2O / 6 moles H2O
Cross-multiplying and solving for x, we have:
x = (1.22 moles NH3 * 6 moles H2O) / 4 moles NH3 = 1.83 moles H2O
Thus, 1.83 moles of gaseous water are produced when 1.22 moles of NH3 are burned.