Answer:
15.35 g of (NH₄)₃PO₄
Explanation:
First we need to look at the chemical reaction:
3 NH₃ + H₃PO₄ → (NH₄)₃PO₄
Now we calculate the number of moles of ammonia (NH₃):
number of moles = mass / molecular wight
number of moles = 5.24 / 17 = 0.308 moles of NH₃
Now from the chemical reaction we devise the following reasoning:
if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄
then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄
X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄
mass = number of moles × molecular wight
mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄
<u>Given:</u>
Mass of MgBr2 = 0.500 g
<u>To determine:</u>
Number of anions in 0.500 g MgBr2
<u>Explanation:</u>
Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol
Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles
Based on stoichiometry-
1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions
Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-
Now,
1 mole of Br- contains 6.023 * 10²³ anions
0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions
Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2
I'm not sure I'm sorry for that but I found this http://www.chem4kids.com/files/atom_ions.html
