The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O 2 ( g ) O2(g) are needed to completely burn 17.9 g C 3 H 8 ( g ) ?

65.1g of O2 will be needed

Explanation:

The equation for the reaction is given below:

C3H8 + 5O2 —>3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of O2 = (16x2) = 32g/mol

Mass conc. of O2 from the balanced equation = 5 x 32 = 160g

From the above calculations, we can easily find the mass of O2 needed to completely burn 17.9 g C3H8. This can be done by the following:

From the equation,

44g of C3H8 was completely burnt by 160g of O2.

Therefore,

17.9g of C3H8 will be burnt by Xg of O2. i.e

Xg of O2 = (17.9x160) / 44 = 65.1g

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

"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"