how many grams of HNO3 can be produced if 225 grams of NO2 is mixed with 55.2 g of H2O?
3NO2 + H2O=2HNO3 + NO
To find the number of grams of HNO3 that can be produced, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reactant, we need to compare the moles of NO2 and H2O. Let's calculate the moles of each reactant using their molar masses:
Molar mass of NO2 (nitrogen dioxide) = 46.01 g/mol
Molar mass of H2O (water) = 18.02 g/mol
Moles of NO2 = 225 g / 46.01 g/mol = 4.89 mol
Moles of H2O = 55.2 g / 18.02 g/mol = 3.06 mol
Now, let's compare the molar ratios between NO2 and HNO3 in the balanced equation:
3 mol NO2 : 2 mol HNO3
From the balanced equation, we can see that 3 moles of NO2 are required to produce 2 moles of HNO3. Therefore, the moles of HNO3 that can be produced will be:
Moles of HNO3 = (2 mol HNO3 / 3 mol NO2) x 4.89 mol NO2 (since NO2 is the limiting reactant)
= 3.26 mol
Now, we can calculate the grams of HNO3 based on its molar mass:
Molar mass of HNO3 (nitric acid) = 63.01 g/mol
Grams of HNO3 = 3.26 mol x 63.01 g/mol = 205.89 g
Therefore, approximately 205.89 grams of HNO3 can be produced.
To find out how many grams of HNO3 can be produced, we need to determine the limiting reactant. The limiting reactant is the one that will be entirely consumed in the reaction and will determine the maximum amount of product that can be formed.
First, we need to convert the masses of NO2 and H2O into moles. To do this, we divide the given mass of each substance by their respective molar masses.
The molar mass of NO2 is 46.01 g/mol, and 225 g of NO2 can be converted to moles as follows:
225 g NO2 * (1 mol NO2 / 46.01 g NO2) = 4.89 mol NO2
The molar mass of H2O is 18.02 g/mol, and 55.2 g of H2O can be converted to moles as follows:
55.2 g H2O * (1 mol H2O / 18.02 g H2O) = 3.06 mol H2O
Next, we need to determine the mole ratio between NO2 and HNO3. From the balanced equation:
3NO2 + H2O -> 2HNO3 + NO
We can see that the ratio between NO2 and HNO3 is 3:2.
To determine the maximum amount of HNO3 that can be produced, we can compare the number of moles of NO2 and H2O and determine the limiting reactant. Since the mole ratio between NO2 and HNO3 is 3:2, we need to compare the number of moles of NO2 to that of HNO3.
Since there are 4.89 moles of NO2 and the mole ratio is 3:2, we can calculate the maximum number of moles of HNO3 as follows:
4.89 mol NO2 * (2 mol HNO3 / 3 mol NO2) = 3.26 mol HNO3
Finally, to convert the moles of HNO3 to grams, we multiply by the molar mass of HNO3, which is 63.01 g/mol:
3.26 mol HNO3 * 63.01 g HNO3 / 1 mol HNO3 = 205.24 g HNO3
Therefore, based on the calculation, the maximum amount of HNO3 that can be produced is 205.24 grams.
205 g of HNO3
Compute the number of moles of NO2 and H2O to determine which reactant is limiting.
There are 55.2/18 = 3.07 moles of H2O and 225/46 = 4.89 moles of NO2. There is more than enough H2O available to react with NO2, so NO2 is the limiting reactant. 4.89 moles of NO2 will resct with 4.89/3 = 1.63 moles of H2O to form 3.26 moles of HNO3. Convert that to grams using the molecular weight of HNO3.