How many molecules and what mass of NO is produced if 138.0g of NO2 is combine with 1.20 e24 molecules of water according to the following reaction: 3NO2+H20--->2HNO3+NO
You have a limiting reagent problem. I know that because amounts for BOTH reactants are given.
Convet 138.0g NO2 to mols. mols = grams/molar mass.
Do the same for 1.20E24 molecules H2O. mol = 1.20E24/6.02E24 = ?
Using the coefficients in the balanced equation, convert mols NO2 to mols NO.
Do the same for mols H2O.
It is likely that the two answers will not be the same which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS tje smaller one and the reagent producing that value is the limiting reagent.
Now convert mols NO to g. g = mols x molar mass.
For # molecules, remember that 1 mol contains 6.02E23 molecules..
To find the number of molecules and the mass of NO produced, we need to use stoichiometry and the given data.
Step 1: Write and balance the equation
The balanced equation is: 3 NO2 + H2O → 2 HNO3 + NO
Step 2: Convert mass of NO2 to moles
The molar mass of NO2 is 46.01 g/mol. To convert grams to moles, divide the given mass by the molar mass:
138.0 g NO2 × (1 mol NO2 / 46.01 g NO2) = 3.0 mol NO2
Step 3: Use stoichiometry to determine the number of moles of NO
From the balanced equation, we find that the stoichiometric ratio between NO2 and NO is 3:1 (3 NO2 molecules produce 1 NO molecule). Therefore, for 3.0 moles of NO2, we will have:
3.0 mol NO2 × (1 mol NO / 3 mol NO2) = 1.0 mol NO
Since 1 mole of any substance contains 6.02 × 10^23 molecules (Avogadro's number), there are 6.02 × 10^23 molecules of NO.
Step 4: Calculate the mass of NO produced
To find the mass of NO, we need to use the molar mass of NO, which is 30.01 g/mol:
1.0 mol NO × (30.01 g NO / 1 mol NO) = 30.01 g NO
Therefore, 138.0 grams of NO2 will produce 6.02 × 10^23 molecules of NO and have a mass of 30.01 grams.