Question

2.67g of manganese(IV) oxide was added to 200.cm3 of 2.00 mol dm-3 HCL. calculate the amount in lol of manganese oxide added

Answer 1

The amount of manganese(IV) oxide added is 0.0307 moles, calculated by dividing the mass of MnO2 (2.67g) by its molar mass (86.94 g/mol).

Step-by-step explanation:

To calculate the amount in moles of manganese(IV) oxide (MnO2) added, you need to use the molar mass of MnO2. The molar mass of MnO2 is 86.94 g/mol. You can find the number of moles by dividing the mass of the substance by its molar mass:

Number of moles

= Mass (g) / Molar mass (g/mol)
In this case, the calculation will be:

Number of moles = 2.67 g / 86.94 g/mol = 0.0307 moles

Therefore, the amount of manganese oxide added is 0.0307 moles.