How many atoms of phosphorus are in 2.30 mol of copper(ii) phosphate? i'm having trouble figuring out the steps. i do know avogadro's number is needed! 6.022x10^23

Answer is 2.77012 x 10²⁴

Explanation;

The formula of copper(II) phosphate is Cu₃(PO₄)₂. Hence, 1 Cu₃(PO₄)₂ has 2 P atoms. The stoichiometric ratio between Cu₃(PO₄)₂ and P atom is 1 : 2.

Hence,

moles of P atoms = moles of Cu₃(PO₄)₂ x 2

= 2.30 mol x 2 = 4.60 mol

According to the Avogadro's constant, 1 mole has 6.022 x 10²³ particles.

Hence, number of P atoms = Avogadro's constant x number of moles

= 6.022 x 10²³ mol⁻¹ x 4.60 mol

= 2.77012 x 10²⁴

i'm pretty sure its b i might be wrong though!

l dont know l am s sorry