Question 2 (3 points) Based on the equation in Question #1, how many liters of Oz at STP can be produced from 12.0 moles of NaClO3?
оа
Ob
Oc
256 L
259 L
403 L
576 L

Volume of oxygen produced =  403 L

Explanation:

Given data:

Number of moles of NaClO₃ = 12.0 mol

Volume of oxygen produced = ?

Temperature = 273 K

Pressure = 1 atm

Solution:

Chemical equation:

2NaClO₃        →     2NaCl + 3O₂

Now we will compare the moles of oxygen and NaClO₃ .

          NaClO₃       :        O₂

               2            ;         3

            12.0          :       3/2×12.0 = 18.0 mol

Volume of oxygen produced:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

1 atm × V = 18.0 mol × 0.0821 atm.L/mol.K × 273 K

V = 403 L .atm / 1 atm

V = 403 L

"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"

the answer is c

tell me if wrong