When nitrogen dioxide (NO2) from car exhaust combines w/ water in the air it forms nitric acid (HNO3), which causes acid rain, and nitrogen monoxide. 3NO2(g)+ H2O(L)----->2HNO3(aq)+ N(g) a.How many molecules of NO2 are needed to react with 0.300 mole H2O?
b. how many grams of HNO3 can be produced if 175g NO2 is mixed with 55.2g of H2O? the first one i posted is mixed up with the other problem from my homework thanks
You made a typo in the equation.
3NO2 + H2O ==> 2HNO3 + NO
0.300 mol H2O x (3 mols NO2/1 mol H2O) - 0.300 x 3/1 = 0.9 mol NO2 needed.
You know 1 mol NO2 has 6.02E23 molecules; you need to calculate the numb err in 0.9 mol.
b. This is a limiting reagent. I work them this way.
mols NO2 = grams/molar mass = ?
mols H2O = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NO2 to mols HNO3.
Do the same to convert mols H2O to mols HNO3.
It is quite likely these two numbers for mols HNO3 will not be the same which means one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent providing that number is the limiting reagent.
Convert the smaller number to grams. g = mols x molar mass.
how will I get the mass of HNO3 produced if 175g NO2 is mixed with 55.2 g H2O. Should I add both the grams of NO2 and H2O
No. The instructions for that part of the problem I gave under part b. Find mold NO2 and mols H2O, convert each to mols HNO3, use the SMALLER number of mols and convert to grams HNO3.
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To solve these types of problems, you need to use the concept of stoichiometry and the balanced chemical equation given. The coefficients in the balanced equation represent the mole ratio between reactants and products.
a. To determine the number of molecules of NO2 needed to react with 0.300 moles of H2O, you need to use the mole ratio between NO2 and H2O from the balanced equation.
From the balanced equation: 3 NO2(g) + H2O(L) → 2 HNO3(aq) + N(g)
The mole ratio between NO2 and H2O is 3:1. This means that for every 3 moles of NO2, 1 mole of H2O is needed.
So, 3 moles of NO2 are needed to react with 1 mole of H2O.
Therefore, if you have 0.300 moles of H2O, you can calculate the number of moles of NO2 needed:
0.300 moles H2O x (3 moles NO2 / 1 mole H2O) = 0.900 moles NO2
Now, to convert moles of NO2 to molecules, you can use Avogadro's number.
1 mole of any substance contains 6.022 x 10^23 molecules.
Therefore, 0.900 moles NO2 = 0.900 x (6.022 x 10^23) molecules = 5.42 x 10^23 molecules of NO2.
Thus, you would need 5.42 x 10^23 molecules of NO2 to react with 0.300 moles of H2O.
b. To determine the number of grams of HNO3 produced when 175 grams of NO2 reacts with 55.2 grams of H2O, you need to use the concept of stoichiometry again.
First, convert the given masses of NO2 and H2O to moles using their respective molar masses:
Molar mass of NO2 = 46.01 g/mol
Molar mass of H2O = 18.02 g/mol
Number of moles of NO2 = 175 g NO2 / 46.01 g/mol = 3.80 moles NO2
Number of moles of H2O = 55.2 g H2O / 18.02 g/mol = 3.06 moles H2O
From the balanced equation: 3 NO2(g) + H2O(L) → 2 HNO3(aq) + N(g)
The mole ratio between NO2 and HNO3 is 3:2. This means that for every 3 moles of NO2, 2 moles of HNO3 are produced.
Using the mole ratio, you can calculate the number of moles of HNO3 produced:
Number of moles of HNO3 = (3.80 moles NO2) x (2 moles HNO3 / 3 moles NO2) = 2.53 moles HNO3
Finally, convert moles of HNO3 to grams using its molar mass:
Molar mass of HNO3 = 63.01 g/mol
Mass of HNO3 = 2.53 moles HNO3 x 63.01 g/mol = 159.84 grams HNO3
Therefore, you can produce 159.84 grams of HNO3 when 175 grams of NO2 is mixed with 55.2 grams of H2O.