Given the reaction represented by the balanced equation:

CH4(g)+3F2(g)->3HF(g)+CHF3 calculate the number of g of CHF3 that could be produced by mixing 180 g F2 with excess CH4.

Explanation:

As given in the question that methane() is taken excess amount

mass of depend only on mass of fluorine

mass of =180 g

mole of = =9.47

3 mole of gives 1 mole of

so 1 mole will give mole of

therefore,

9.47 mole of will give mole of

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3