4 NH3 + 5 O2 > 4 NO + 6 H2O
How many moles and how many grams of oxygen (O2) are needed to react with 56.8 grams
of ammonia by this reaction?

The moles and mass of oxygen gas needed to react is 4.175 moles and 133.6 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of ammonia = 56.8 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

4 moles of ammonia reacts with 5 mole of oxygen gas

So, 3.34 moles of ammonia will react with = of oxygen gas

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 4.175 moles

Putting values in equation 1, we get:

Hence, the moles and mass of oxygen gas needed to react is 4.175 moles and 133.6 grams

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

14,16 18 reacts with group