Question

calculate the mass and volume of oxygen at STP to convert 54 grams of graphite into carbon dioxide​

Answer 1

Explanation:

The equation for the above reaction is as follows:

(graphite is an allotrope of carbon--C)

       C         +       O2 ------->  CO2

         (s)                  (g)                (g)

Therefore 1 mole of graphite requires 1 Mole of O2 to give 1 mole of CO2

Calculate the moles of graphite:

molar mass of C = 12

mass = 2400 g 

moles = mass/molar mass

           = 2400/12

           =200 moles

if 1 mole of C requires 1 mole of O2

then 200 moles requires  = 1 x 200 / 1

                                         = 200 moles

  

            Find mass of oxygen: (molar mass of O2= 32, moles=200)

mass = moles x molar mass

         = 200 x 32

         = 6400g

          = 6.4kg of oxygen

Find the volume of oxygen:

Ideal gas law..... states that 1 mole of an ideal gas occupies 22.4 litres.

                       

                     therefore 200 moles of O2 occupies= (22.4 x 200) liters

   

                                                                               =4480 liters

                 volume of oxygen = 200

Answer 2

Explanation:

C + O2 --> CO2

given wt. of carbon= 54g

Mol. Wt O2 = 32g

at. wt. C = 12g

12g C --> 32g O2

1g C --> 32/12g O2

54g C --> 32/12 x 54 =144g O2

so at NTP

moles = vol./22.4

wt./mol.wt = vol./22.4

144/32 = vol./22.4

vol. of O2 = 100.8L

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