Ametallurgist begins with 1,250 grams of vanadinite (pb5(vo4)3cl) and ends up with 135 grams of pure vanadium and 915 grams of pure lead. the percent composition of vanadium in the ore is , and the percent composition of lead is

1. 10.8%

2. 73.2%

Sample: 1,250 grams of vanadinite (Pb₅ (VO₄)₃Cl

Mass of Vanadium = 135 g

Mass of Pb = 915 g

Percentage of V = ?

Percentage of Pb =?

Using the formula;

The percentage composition is given by the formula;

= ( mass in grams of element / mass in grams of sample) × 100

Therefore;

The percentage of Vanadium;

  = (mass of V / mass of sample) × 100 = (135 g/ 1,250 g) × 100

   = 10.8 %

The percentage composition of Lead;

 = (mass of Pb / mass of sample) × 100 = (915 g / 1,250 g) × 100

  = 73.2%

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 !