How many grams of calcium phosphate can be produced when 89.3 grams of calcium chloride reacts with excess sodium phosphate?

unbalanced equation: cacl2 + na3po4 → nacl + ca3(po4)2

show, or explain, all of your work along with the final answer.

 The grams  of calcium phosphate  that can be  produced when  89.3  grams  of calcium chloride  reacted  with   excess  sodium phosphate is

     83.08  grams  of calcium phosphate

   calculation

Step 1: write a balanced chemical equation

3CaCl₂  + 2Na₃Po₄ →6NaCl + Ca₃(PO₄)₂

step 2: find the moles of CaCl₂

moles  = mass÷  molar mass

from periodic table the molar mass  of CaCl₂ = 40 +( 2 x 35.5) =111 g/mol

moles = 89.3 g÷111 g/mol = 0.805  moles

Step 3: use the  mole ratio to determine the moles of Ca₃(PO₄)₂

CaCl₂: Ca₃(PO₄)₂  is  3:1 therefore the  moles of Ca₃(PO₄)₂ =0.805  x 1/3 =0.268  moles

Step 4: find mass  of Ca₃(PO₄)₂

mass = moles x  molar mass

from periodic table the molar mass of Ca₃(PO₄)₂ = (40 x3) +[ 31 +(16 x4)]2 =310 g/mol

mass is therefore =0.268  moles x 310 g/mol =83.08 grams

h let me think

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2