Question

In a different experiment, Kai uses 1.00 g of ethene (C2H4) in the reaction. What volume,in ml, of the 0.200 M KMnO4 reacts with the

ethene completely?
3C2H4 (1) + 2KMnO4(aq) + 4H2O(1)
3C2H6O2(1) + 2MnO2(s) + 2KOH(aq)-mL KMnO4

Answer 1

To find the volume of 0.200 M KMnO4 that will completely react with 1.00 g of ethene (C2H4), we first calculate the moles of ethene, use the stoichiometry of the reaction to determine the moles of KMnO4 needed, and then the volume based on the molarity. The calculation shows that 118.5 mL of 0.200 M KMnO4 is required.

Step-by-step explanation:

The volume of 0.200 M KMnO4 required to react with 1.00 g of ethene (C2H4) can be determined using stoichiometry and the molar mass of ethene. First, calculate the number of moles of C2H4 in 1.00 g:

Number of moles = mass (g) ÷ molar mass (g/mol)

For C2H4, molar mass is approximately 28.05 g/mol:

Number of moles of C2H4 = 1.00 g ÷ 28.05 g/mol = 0.0356 moles

From the balanced chemical equation, 3 moles of C2H4 react with 2 moles of KMnO4:

Number of moles of KMnO4 required = (2 moles KMnO4 ÷ 0.0356 moles C2H4) ÷ 3 moles C2H4

Number of moles of KMnO4 required = 0.0237 moles

To find the volume of 0.200 M KMnO4:

Volume (L) = moles ÷ molarity

Volume of KMnO4 = 0.0237 moles ÷ (1 L/0.200 moles)

Volume of KMnO4 = 0.1185 L = 118.5 mL

Therefore, 118.5 mL of 0.200 M KMnO4 is required to completely react with 1.00 g of ethene.