Question

Calculate the # of moles present.
1.34kg of potassium bromide

Answer 1

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

Answer 2

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

Answer 3

@.Sam. Well done :)