## Chiomatn93 Group Title Calculate the # of moles present. 1.34kg of potassium bromide 2 years ago 2 years ago

1. oOKawaiiOo

1.34 kg --- 0.00134 grams/ MW of KBR = # of moles

2. .Sam.

You can use the ratio method, For 1 mole of KBr you have 39+80 g of KBr $1340 \text{g of KBr} \times \frac{1~\text{mole of KBr}}{39+80~\text{g of KBr}}$ $1340 \cancel{\text{g of KBr}} \times \frac{1~\text{mole of KBr}}{39+80~\cancel{\text{g of KBr}}}=11.26 \text{ moles of KBr}$

3. oOKawaiiOo

@.Sam. Well done :)