Question

What volume of 0.78mol/L KIO3(aq) with react with 4.8g of Br2 (aq).

Answer 1

To calculate the volume of KIO3 (aq) that will react with 4.8 g of Br2 (aq), we need to use the balanced chemical equation for the reaction between KIO3 and Br2.

The balanced chemical equation is:

2 KIO3 + 10 Br2 → 10 KBr + 2 I2 + 6 O2

From the balanced equation, we can see that the mole ratio between KIO3 and Br2 is 2:10, meaning that 2 moles of KIO3 will react with 10 moles of Br2.

First, we need to calculate the number of moles of Br2 in 4.8 g.

The molar mass of Br2 is 159.808 g/mol.

Number of moles of Br2 = Mass of Br2 / Molar mass of Br2
Number of moles of Br2 = 4.8 g / 159.808 g/mol
Number of moles of Br2 = 0.03 mol

Since the mole ratio between KIO3 and Br2 is 2:10, the number of moles of KIO3 needed can be calculated using this ratio.

Number of moles of KIO3 = Number of moles of Br2 × (2 moles of KIO3 / 10 moles of Br2)
Number of moles of KIO3 = 0.03 mol × (2/10)
Number of moles of KIO3 = 0.006 mol

Finally, we can calculate the volume of KIO3 using the concentration of KIO3.

Concentration of KIO3 = 0.78 mol/L
Volume of KIO3 = Number of moles of KIO3 / Concentration of KIO3
Volume of KIO3 = 0.006 mol / 0.78 mol/L
Volume of KIO3 = 0.0077 L or 7.7 mL

Therefore, 7.7 mL of 0.78 mol/L KIO3 (aq) will react with 4.8 g of Br2 (aq).