4. Calculate the number of liters of oxygen gas needed to produce 15.0 liters of dinitrogen trioxide. Assume all gases are at the same conditions of temperature and pressure.
2N2(g) + 3O2(g)  2N2O3(g)

22.5L

Explanation:

Given parameters:

Volume of N₂O₃  = 15liters

Unknown:

volume of oxygen gas  = ?

Solution:

To solve this problem, we need to work from the known to unknown. The known is N₂O₃. We can find the number of moles from here.

Note: The condition is Standard temperature and pressure

   Number of moles  =

    Number of moles of N₂O₃ =   = 0.67moles

The equation of the reaction;

                            2N₂  +  3O₂   →   2N₂O₃

Since the equation is balanced;

                      2 moles of N₂O₃ is produced from 3 moles of O₂

                  0.67moles  of N₂O₃ is produced from   = 1.01 moles of O₂

Therefore;

               Volume of O₂   = 1.01 x 22.4  = 22.5L

beta decay is the answer

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"