3NO2 + H2O = 2HNO3 + NO
How many grams of nitrogen dioxide are required to produce 5.89 x 10 to the 3rd power kg HNO3 in excess water?
N2O5 + H2O ==> 2HNO3
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Well, let's do some calculations. Since we have the balanced equation 3NO2 + H2O = 2HNO3 + NO, we know that for every 3 moles of NO2, we produce 2 moles of HNO3.
To convert 5.89 x 10^3 kg of HNO3 to grams, we multiply by 1000 to get 5.89 x 10^6 grams.
Now, we need to convert grams of HNO3 to moles using the molar mass of HNO3:
1 mole of HNO3 = 63 grams
So, 5.89 x 10^6 grams of HNO3 is equal to (5.89 x 10^6) / 63 = 9.34 x 10^4 moles of HNO3.
Since the ratio between NO2 and HNO3 is 3:2, we can say that for every 2 moles of HNO3 produced, we need 3 moles of NO2. Therefore, we would need (9.34 x 10^4) x (3/2) = 1.40 x 10^5 moles of NO2.
Now, to convert moles back to grams, we need to know the molar mass of nitrogen dioxide (NO2):
1 mole of NO2 = 46 grams
So, 1.40 x 10^5 moles of NO2 is equal to (1.40 x 10^5) x 46 = 6.44 x 10^6 grams of NO2.
Therefore, you would need approximately 6.44 x 10^6 grams of nitrogen dioxide to produce 5.89 x 10^3 kg of HNO3 in excess water.
Now that's a lot of NO2! I hope you're not planning on blowing up any balloons with that amount!
To solve this problem, we first need to calculate the molar mass of nitrogen dioxide (NO2) and nitric acid (HNO3).
The molar mass of NO2 is:
N = 14.01 g/mol
O = 16.00 g/mol (x2 since there are 2 oxygen atoms)
Total molar mass of NO2 = 14.01 + (16.00 x 2) = 46.01 g/mol
The molar mass of HNO3 is:
H = 1.01 g/mol
N = 14.01 g/mol
O = 16.00 g/mol (x3 since there are 3 oxygen atoms)
Total molar mass of HNO3 = 1.01 + 14.01 + (16.00 x 3) = 63.01 g/mol
Now, we can set up a ratio using the balanced chemical equation:
3 NO2 + H2O → 2 HNO3 + NO
From the equation, we can see that 3 moles of NO2 produce 2 moles of HNO3. So, the ratio is:
3 mol NO2 / 2 mol HNO3
Convert the given mass of HNO3 to moles:
5.89 x 10^3 kg HNO3 = 5.89 x 10^3 x 10^3 g HNO3 = 5.89 x 10^6 g HNO3
Moles of HNO3 = mass / molar mass = 5.89 x 10^6 g / 63.01 g/mol
Now, to find the moles of NO2 required, we use the ratio from the balanced equation:
Moles of NO2 = Moles of HNO3 x (3 mol NO2 / 2 mol HNO3) = (5.89 x 10^6 g / 63.01 g/mol) x (3 mol NO2 / 2 mol HNO3)
Finally, we can convert moles of NO2 to grams:
Grams of NO2 = Moles of NO2 x molar mass of NO2 = (5.89 x 10^6 g / 63.01 g/mol) x 46.01 g/mol
Now, we can perform the calculations to find the value of grams of NO2 required to produce 5.89 x 10^3 kg of HNO3.
To determine the number of grams of nitrogen dioxide (NO2) required to produce 5.89 x 10^3 kg of nitric acid (HNO3), we need to use stoichiometry and convert from kg to grams.
First, let's balance the chemical equation:
3NO2 + H2O -> 2HNO3 + NO
The balanced equation tells us that for every 3 moles of nitrogen dioxide (NO2), we obtain 2 moles of nitric acid (HNO3). So, we need to convert the given mass of HNO3 into moles.
1 mole of HNO3 has a molar mass of 63 g/mol (atomic masses of hydrogen, nitrogen, and oxygen combined).
5.89 x 10^3 kg HNO3 can be converted to grams by multiplying by 1000 since there are 1000 grams in a kilogram:
5.89 x 10^3 kg HNO3 = 5.89 x 10^6 g HNO3
Now, we can calculate the number of moles of HNO3 using its molar mass:
moles of HNO3 = (mass of HNO3 / molar mass of HNO3)
moles of HNO3 = (5.89 x 10^6 g HNO3 / 63 g/mol)
Next, we use the stoichiometry of the balanced equation to find the number of moles of NO2 required to produce the calculated moles of HNO3.
According to the balanced equation, 3 moles of NO2 produce 2 moles of HNO3. Therefore:
moles of NO2 = (moles of HNO3) x (3 moles of NO2 / 2 moles of HNO3)
Finally, we convert the moles of NO2 into grams by multiplying by the molar mass of NO2, which is 46 g/mol:
mass of NO2 (in grams) = (moles of NO2) x (molar mass of NO2)
Now, we can plug in the previously calculated values:
mass of NO2 = (moles of NO2) x (46 g/mol)
mass of NO2 = [(moles of HNO3) x (3 moles of NO2 / 2 moles of HNO3)] x (46 g/mol)
By substituting the calculated values into this equation, you can determine the mass of NO2 required to produce the given amount of HNO3 in excess water.