Question

Na2co3 + ca(no3)2 --> 2nano3 + caco3 if the problem tells you that you have 45 g of sodium carbonate and 35 g of calcium nitrate, a) which is the limiting reagent and b) what is the theoretical amount of calcium carbonate that will precipitate?

Answer 1

Calcium nitrate is the limiting reagent and theoretical yield of calcium carbonate is 21.3 grams.

What is the relation between mass & moles?

Relation between the mass and moles of any substance will be represented as:

• n = W/M, where
• W = given mass
• M = molar mass

Moles of Na₂CO₃ = 45g / 105.98g/mol = 0.424 mol

Moles of Ca(NO₃)₂ = 35g / 164g/mol = 0.213 mol

From the stoichiometry of the reaction it is clear that:

• 1 mole of Na₂CO₃ = reacts with 1 mole of Ca(NO₃)₂
• 0.424 mole of Na₂CO₃ = reacts with 0.424 mole of Ca(NO₃)₂

It means Ca(NO₃)₂ is the limiting reagents as less moles of this is present in the reaction, so the formation of product is depends on the amount of Ca(NO₃)₂.

Again from the stoichiometry of the reaction, it is clear that:

• 1 mole of Ca(NO₃)₂ = produces 1 mole of CaCO₃
• 0.213 mole of Ca(NO₃)₂ = produces 0.213 mole of CaCO₃

Theoretical mass of CaCO₃ = (0.213mol)(100g/mol) = 21.3g

Hence, required mass of calcium carbonate is 21.3g.

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