Answer to Question #234238 in Chemistry for Henri

Question

  295 gram Rubidium Oxide reacted with 202 gram calcium cyanide to produce Rubidium cyanide  and Calcium  Oxide.  Find

a)gram  of  Calcium  oxide

b) gram of excess  reactant  used up

c) gram of  excess  reactant that remained  unreacted

Answer 1

Solution:

Calculate moles of each reactant:

(295 g Rb₂O) × (1 mol Rb₂O / 186.94 g Rb₂O) = 1.578 mol Rb₂O

(202 g Ca(CN)₂) × (1 mol Ca(CN)₂ / 92.1128 g Ca(CN)₂) = 2.193 mol Ca(CN)₂

Balanced chemical equation:

Rb₂O + Ca(CN)₂ → 2RbCN + CaO

According to stoichiometry:

1 mol of Rb₂O reacts with 1 mol of Ca(CN)₂.

Thus, 1.578 mol of Rb₂O reacts with:

(1.578 mol Rb₂O) × (1 mol Ca(CN)₂ / 1 mol Rb₂O) = 1.578 mol Ca(CN)₂

However, initially there is 2.193 mol of Ca(CN)₂ (according to the task).

Therefore, Rb₂O acts as limiting reagent and Ca(CN)₂ is excess reactant.

According to stoichiometry:

1 mol of Rb₂O produces 1 mol of CaO.

Thus, 1.578 mol of Rb₂O produces:

(1.578 mol Rb₂O) × (1 mol CaO / 1 mol Rb₂O) = 1.578 mol CaO

Calculate the mass of calcium oxide (CaO):

(1.578 mol CaO) × (56.0774 g CaO / 1 mol CaO) = 88.49 g CaO = 88.5 g CaO

Calculate the mass of excess reactant used up (Ca(CN)₂):

(1.578 mol Ca(CN)₂) × (92.1128 g Ca(CN)₂ / 1 mol Ca(CN)₂) = 145.35 g Ca(CN)₂

Calculate the mass of excess reactant that remained unreacted (Ca(CN)₂):

202 g Ca(CN)₂ - 145.35 g Ca(CN)₂ = 56.65 g Ca(CN)₂

Answers:

a) 88.5 g CaO

b) 145.35 g Ca(CN)₂

c) 56.65 g Ca(CN)₂