Question

If 0.86 mole of MnO2 and 48.2 g of HCL react, which reagent will be used up first? How many grams of Cl2 will be produced? Calculate the number of grams of the excess reagent remaining at the end of the reaction

Answer 1

Amount of Cl2 produced = 23 g

Amount of excess reagent MnO2 remaining = 46 g

Step-by-step explanation:

The reaction between MnO2 and HCl forming MnCl2, Cl2 and H2O can be expressed as:

`4HCl + MnO2 \rightarrow Cl2 + MnCl2 + 2H2O`

The amount of Cl2 produced will be determined by the amount of the limiting reactant.

Moles of MnO2 = 0.86 moles

`Moles\ HCl = (Mass\ HCl)/(Mol.wt.\ HCl) = (48.2g)/(36.46g/mol) =1.32moles`

The mole ratio between HCl:MnO2 = 4:1

If all 1.32 moles of HCl were used up then based on the mole ratio, the moles of MnO2 that would be needed is:

`Moles\ MnO2\ needed = (1)/(4) *moles\ of\ HCl\\\\= (1)/(4) *1.32 = 0.33 moles`

However, there are 0.86 moles of MnO2 i.e. there is an excess of MnO2. Hence HCl is the limiting reactant.

Based on the reaction stoichiometry:

4 moles of HCl produces 1 mole of Cl2

`Moles\ Cl2\ produced = (1)/(4) *moles\ of\ HCl\\\\= (1)/(4) *1.32 = 0.33 moles`

`Mass\ Cl2 = moles*mol wt = 0.33moles*70.9g/mol = 23 g`

Excess reagent = MnO2

Moles of MnO2 remaining = 0.86 -0.33=0.53 moles

`Mass\ MnO2\ remaining = moles*mol wt = 0.53moles*86.9g/mol = 46g`