Question

Calculate how many grams of product are formed when 24g of Cs reacts with 39g of Cl₂? (Write and balance the equation)

2Cs + Cl2 → 2CscI Cs = 132.91g/mol Cl = 35.45 g/mol mass ncs 24gCs T = 0.18057 mol Cs Molar Mass mol mass nc2=- Molar Mass 39g Cl2 +2 70.906 gCl2 = 0.550 mol Cl2 I got this wrong: 2 mol cs 24965 x - X 2 mol Esel 2 mol cs 168.365 g CsCl = 15.201qCsCl (this is the wrong answer) 2 mol CsCl 265.82-gcs

Answer 1

molar mass Cs = 132.91 g/mol

molar mass Cl = 35.45 g/mol

molar mass CsCl = 168.36 g/mol

Balanced reaction : 2 Cs + Cl₂ $\rightarrow$ 2 CsCl

Case 1 : Cs is the limiting reactant

mass Cs = 24 g

mass CsCl formed = (mass Cs) * (1 mol Cs / 132.91 g Cs) * (2 mol CsCl / 2 mol Cs) * (168.36 g CsCl / 1 mol CsCl)

mass CsCl formed = (24 g Cs) * (1 mol Cs / 132.91 g Cs) * (2 mol CsCl / 2 mol Cs) * (168.36 g CsCl / 1 mol CsCl)

mass CsCl formed = (24 * 168.36 / 132.91) g CsC

mass CsCl formed = 30.4 g

Case 2 : Cl₂ is the limiting reactant

mass Cl₂ = 39 g

mass CsCl formed = (mass Cl₂) * (1 mol Cl₂ / 70.90 g Cl₂) * (2 mol CsCl / 1 mol Cl₂) * (168.36 g CsCl / 1 mol CsCl)

mass CsCl formed = (39 g Cl₂) * (1 mol Cl₂ / 70.90 g Cl₂) * (2 mol CsCl / 1 mol Cl₂) * (168.36 g CsCl / 1 mol CsCl)

mass CsCl formed = (39 * 2 * 168.36 / 70.90) g CsCl

mass CsCl formed = 185.2 g

Since less mass of CsCl is formed in Case 1, therefore, Cs is the limiting reactant.

mass CsCl formed = 30.4 g