Consider a situation in which 211 g
g
of P4
P
4
are exposed to 240 g
g
of O2
O
2
.What is the maximum amount in moles of P2O5
P
2
O
5
that can theoretically be made from 211 g
g
of P4
P
4
and excess oxygen?
Express your answer to three significant figures and include the appropriate units.

3.00 mol

Explanation:

Given data:

Mass of P₄ = 211 g

Mass of oxygen = 240 g

Moles of P₂O₅ = ?

Solution:

Chemical equation:

P₄ + 5O₂       →     2P₂O₅

Number of moles of P₄:

Number of moles = mass/ molar mass

Number of moles = 211 g / 123.88 g/mol

Number of moles = 1.7 mol

Number of moles of O₂ :

Number of moles = mass/ molar mass

Number of moles = 240 g / 32g/mol

Number of moles = 7.5 mol

Now we will compare the moles of product with reactant.

                       O₂         :         P₂O₅

                        5          :           2

                        7.5       :        2/5×7.5 = 3.00

                       P₄          :         P₂O₅

                        1           :           2

                       1.7         :       2×1.7 = 3.4 mol

Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.

Mass of P₂O₅:

Mass = number of moles × molar mass

Mass = 3 mol ×283.9 g/mol

Mass = 852 g

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"