4 NH3 + 5 O2 --> 4 NO + 6 H2O.
How many grams of O2 are required to product 90 g of NO?
Please assist in solving.
Convert 90 g NO to mols NO. #mols = grams/molar mass.
Convert mols NO to mols O2 using the coefficients in the balanced equation.
Convert mols O2 to grams O2.
grams O2 = mols O2 x molar mass O2.
Still lost. How do you convert mols NO to mols O2?
Use the coefficients in the balanced equation. It's just the usual dimensional kind of problem.
mols O2 = mols NO x (5 mols O2/4 mols NO) = ??
Note that the numbers come from the coefficients. Note that we know which material to put on the top and bottom by choosing those that will make the unit come out right. For example, see the mols NO cancel (that's the unit we DOM'T want) and it leaves mols O2 for the unit (and that's the unit we DO want to convert to). So all that conversion does is to convert from mols O2 to mols NO.
Put in brief form, it is
mols O2 = mols NO x factor and there are only two ways to write the factor. One way is as above (5 mols O2/4 mols NO) and the other way is 4 mols NO/5 mols O2). So how do you know NOT to write
mols O2 = mols NO x (4 mols NO/5 mols O2)?? Because the units won't cancel. In fact, the unit would NOT be mols O2 (which we want) but mols NO^2/mols O2.
I hope this helps.
76.8
To determine how many grams of O2 are required to produce 90 g of NO in the given reaction, we need to use stoichiometry.
Stoichiometry is the calculation of relative quantities of reactants and products in a chemical reaction. In this case, we can use the balanced equation to compare the molar ratios of NH3, O2, NO, and H2O. The balanced equation is:
4 NH3 + 5 O2 -> 4 NO + 6 H2O
From the equation, we can gather the following information (which represents the molar ratios):
- 4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO
- Therefore, the ratio of moles of NH3 to O2 is 4:5
Now, let's calculate the molar mass of O2:
Oxygen (O) atomic mass = 16 g/mol
Molar mass of O2 = 16 g/mol x 2 = 32 g/mol (since O2 contains two oxygen atoms)
To find the number of moles of O2 required to produce 90 g of NO, we can use the molar mass and the molar ratio:
Number of moles of O2 = (Mass of NO / Molar mass of NO) x (Molar ratio of O2 to NO)
The molar mass of NO = 14 g/mol + 16 g/mol = 30 g/mol (adding the atomic masses of nitrogen and oxygen in NO)
Now we can substitute the values into the equation:
Number of moles of O2 = (90 g / 30 g/mol) x (5 moles O2 / 4 moles NO)
Number of moles of O2 = 3 moles O2
Finally, we can calculate the mass of O2 using its molar mass:
Mass of O2 = Number of moles of O2 x Molar mass of O2
Mass of O2 = 3 moles x 32 g/mol = 96 g
Therefore, 96 grams of O2 are required to produce 90 grams of NO.