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
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.
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