9 How many grams of O₂ are needed to produce 15.5 g Fe₂O3 in the following reaction? Fe(s) + O₂(g) → Fe₂O3 (s)​

Explanation:

so u can work out the amount of moles in FeO3 by doing mr of fe3o3 is 55.8*3+16*3=215.4

moles= mass/mr so you do 15.5g/215.4=0.0719 moles

then using 1 to 1 ratio so O2 moles is 0.0719

then use the equation mass=mole*mr

so 0.0719*16=1.15g

hope this make sense :)

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