If 1.0g of rubidium and 1.0 g of bromine are reacted what will be left in measurable amounts in the reaction

0.0639 g of Br₂ will be left

Explanation:

The balance chemical equation for given synthesis reaction is,

2 Rb + Br₂ → 2 RbBr

Step 1: Calculate Moles of each reactant for given masses as;

Moles of Rb = 1 g : 85.46 g/mol

Moles of Rb = 0.0117 moles

Similarly,

Moles of Br₂ = 1 g : 159.80 g/mol

Moles of Br₂ = 0.00625 moles

Step 2: Find reactant in Excess as:

According to equation,

2 moles of Rb reacted with = 1 mole of Br₂

So,

0.0117 moles of Rb will react with = X moles of Br₂

Solving for X,

X = 0.0117 mol * 1 mol / 2 mol

X = 0.00585 mol of Br₂

While, as calculated above we are provided with 0.00625 moles of Br₂. Therefore, Br₂ is in excess and the excess amount is calculated as,

= Given Moles - Consumed Moles

= 0.00625 moles - 0.00585 moles

= 0.0004 moles

Step 3: Calculate mass of excess Br₂ as:

Mass = Moles * M. Mass

Mass = 0.0004 mol * 159.80

Mass = 0.0639 g of Br₂

Note: Speaking broadly, there will be 1.93 g of RbBr along with 0.0639 g of Br₂ because RbBr is being produced and is available along with unreacted Br₂.