What is the maximum amount of p4 that can be produced from 1.0 kg of phosphorite?

Answer and explanation;

-Phosphorous can be prepared from calcium phosphate by the following reaction:   2Ca3(PO4)2 + 6SiO2 + 10C > 6CaSi03 + P4 + 10CO

Phosphorite is a mineral that contains Ca3(PO4)2 plus other non phosphorous containing compounds.

-If the phosphorite sample is 75% Ca3(PO4)2 by mass, then;

mass of Ca3(PO4)2 is given by; 0.75 * (1000 g) = 750g

MM Ca3(PO4)2 = 310.19 g/mol  

thus, Number of moles of  Ca3(PO4)2;

   = 750g / 310.19 g/mol

    = 2.42 moles of Ca3(PO4)2   From the equation;

(2 mol Ca3(PO4)2 = 1 mol P4)   Thus; moles of P4 will be;

=2.42 mol Ca3(PO4)2 * (1 mol P4 / 2 mol Ca3(PO4)2)

= 1.21 moles of  P4   Mass of P4 will be;

=1.21 mol P4 * ( 4 x 30.974) g/mol

= 1.21 mol P4 * 123.896 g/mol

= 150 g P4.  (2 sf)

Hence;

1.5 x10^(2) g P4 are produced from 1.0 kg Phosphorite mineral containing 75% Ca3(PO4)2.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

c- cold periods in earth's history during which glaciers covered much of the surface.

explanation: