# In the reaction in a closed vessel of calcium with oxygen, taken in the same mass ratios, 112 g of CaO was obtained

In the reaction in a closed vessel of calcium with oxygen, taken in the same mass ratios, 112 g of CaO was obtained. What substance and how many grams of it remained in excess after the reaction Decision:
M (CaO) = 56 g / mol;
M (O2) = 32 g / mol.
The reaction equation has the form:
2Ca + O2 = 2CaO
From the reaction equation it follows that to obtain 2 mol of calcium oxide, 2 mol of metallic calcium and 1 mol of oxygen are required.
Then the amount of calcium oxide obtained is equal to, we get:
n (CaO) = m (CaO) / M (CaO) = 112 g / 56 g / mol = 2 mol
So, to obtain 112 g (2 mol) of calcium oxide, 80 g (2 mol) of calcium metal and 32 g (1 mol) of oxygen were taken. Since the starting materials were taken in equal mass ratios, oxygen was taken in excess.
Calculates the remaining oxygen in the mixture, we get:
residue (O2) = m (Ca) – m (O2) = 80 g – 32 g = 48 g.
Answer: residual (O2) = 48 g. Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.