Question

Calculate the mass of 4.50 moles of Ca3(PO3)4

Answer 1

Stoichiometry:
First, calculate the number of grams for one mole of Ca3 (PO3)4
(3 * (Mass of Ca)) + (4 * (Mass of P + (3 * Mass of Oxygen)))
= (3*40.08) + 4(30.97 + (3*16.00))
=(120.24) + 4(78.97)
=436.12 g / mol Ca3(PO3)4
This means there are 436.12 g per 1 mole of Ca(PO3)4. Since there are 4.50 moles of Calcium Phosphate, mulitply the molar mass of Ca(PO3)4 by 4.50 and you should get 1962.54 g. Since there are 3 sigfigs, the final answer is 1960 g.

on a side note: I put in all my work in case 1. your periodic table if different, 2. my work is wrong, 3. you put in the question wrong because I feel that the actual compound would be Ca3(PO4)3 instead of Ca3(PO3)4 (if this is the case, the answer should be 1820 g).

Answer 2

Answer: 1962.513u

Step-by-step explanation:

Ca3(PO3)4

Ca = 40.078u

P = 30.97u

O = 16u

1 mole of Ca3(PO3)4 =(3*40.078) + 4[(16*3) + 30.97]

1 mole = 120.234 + 315.88 = 436.114g/mol

If 1 mole = 436.114g/mole

4.5 mole = x?

Cross multiply and make x the subject,

x = 4.5*436.114/1

= 1962.513g/mol

Therefore, 4.5 moles of Ca(PO3)4 equals 1962.513g/mole