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Answer:
18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
Explanation:
Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.
The decomposition of mercury oxide is given by the chemical equation below:
2HgO ----> 2Hg + O₂
2 moles of HgO decomposes to produce 1 mole of Hg
2 moles of HgO has a mass of 433.2 g
433.2 g of HgO produces 216.6 g of Hg
18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg
Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
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
