How many grams of Nitric oxide are required to produce 145g of nitrogen in this reaction:
Al2O3(s)+6NO(g)---->5N2(g)+6H2o(l)
Follow the steps in this worked example.
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To calculate the grams of Nitric oxide (NO) required to produce 145g of nitrogen (N2), we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is:
Al2O3(s) + 6NO(g) → 5N2(g) + 6H2O(l)
The mole ratio between NO and N2 is 6:5.
This means that for every 6 moles of NO, we get 5 moles of N2.
To find the number of moles of N2, we divide the given mass of nitrogen by its molar mass.
Molar mass of N2 = 28.02 g/mol (2 x 14.01 g/mol)
Number of moles of N2 = 145g ÷ 28.02 g/mol ≈ 5.18 mol
From the balanced equation, we know that the mole ratio between NO and N2 is 6:5.
So, for every 5 moles of N2, we need 6 moles of NO.
Number of moles of NO = (5.18 mol NO/5 mol N2) x (6 mol NO/1 mol N2) ≈ 6.22 mol NO
Finally, to find the mass of NO in grams, we multiply the number of moles of NO by its molar mass.
Molar mass of NO = 30.01 g/mol (14.01 g/mol + 16.00 g/mol)
Mass of NO = 6.22 mol NO x 30.01 g/mol ≈ 186.4 g of NO
Therefore, approximately 186.4 grams of Nitric oxide (NO) are required to produce 145g of nitrogen (N2) in this reaction.
To find out how many grams of Nitric oxide (NO) are required to produce 145g of nitrogen (N2), we need to use stoichiometry.
1. Begin by writing down the balanced chemical equation, including the coefficients:
Al2O3(s) + 6NO(g) → 5N2(g) + 6H2O(l)
2. Determine the molar masses of the substances involved:
- Molar mass of NO = atomic mass of N + atomic mass of O = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
- Molar mass of N2 = atomic mass of N × 2 = 14.01 g/mol × 2 = 28.02 g/mol
3. Establish the molar ratio between Nitric oxide (NO) and nitrogen (N2) by comparing their coefficients in the balanced equation:
From the balanced equation, we can see that 6 moles of NO produce 5 moles of N2.
4. Calculate the number of moles of nitrogen (N2) required:
To find the number of moles of N2 that can be produced from 145g, divide the mass by the molar mass of N2:
Moles of N2 = mass of N2 / molar mass of N2 = 145g / 28.02 g/mol ≈ 5.176 mol
5. Apply the molar ratio to determine the number of moles of Nitric oxide (NO) required:
Since the molar ratio between NO and N2 is 6:5, we can set up a proportion:
Moles of NO / Moles of N2 = 6 / 5
Let x be the number of moles of NO:
x / 5.176 mol = 6 / 5
Cross-multiply and solve for x:
5x = 6 × 5.176
x ≈ 6.211 mol
6. Calculate the mass of Nitric oxide (NO) required:
To find the mass of NO required, multiply the number of moles of NO by its molar mass:
Mass of NO = moles of NO × molar mass of NO = 6.211 mol × 30.01 g/mol ≈ 186.43 g
Therefore, approximately 186.43 grams of Nitric oxide (NO) are required to produce 145 grams of nitrogen (N2) in the given reaction.