Question

Part B 2K(s) + Br2(l) → 2 KBr(s) Express your answer using four significant figures.

Answer 1

15.83 g of Br2 (molar mass 159.808 g/mol) corresponds to

$\frac{\text { 15.83 g }}{\text { 159.8 g/mol}}=\text { 0.09906 mol}$

1 mole of Br2 gives 2 moles of KBr.

0.09906 moles of Br2 gives $2 \times \text { 0.09906 mol}=\text { 0.198123 mol}$ of KBr.

The molar mass of KBr is 119 g/mol.

Mass of KBr formed $=\frac{\text { 0.198123 mol}}{ \text { 119 g/mol }} = \text { 0.001665 g}$