2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
Cc stands for cm cubed (cubic centimetre).
So
6cc=6 cm^3.
Nickel is a pure substance
Rust is also a pure substance
The heat absorbed by the water is
Q = 500 (4.18) (32.2 - 25)
Q = 15048 J
The enthalpy of fusion of the sodium acetate is:
<span>ΔHf = Q / m
</span><span>ΔHf = 15048 / 100
</span>ΔHf = 150.48 J/g
8. 4 grams of water will be produced.
