Question

Basic solutions of Na₄XeO₆ are powerful oxidants. What mass of Mn(NO₃)₂•6H₂O reacts with 125.0 mL of a 0.1717 M basic solution of Na₄XeO₆ that contains an excess of sodium hydroxide if the products include Xe and a solution of sodium permanganate?

a. 3.27 g
b. 4.18 g
c. 5.94 g
d. 6.55 g

Answer 1

d. 6.55 g

The mass of Mn(NO3)2 6H2O that reacts with 125.0 mL of a 0.1717 M basic solution of Na4XeO6 is 7.21 g.

Step-by-step explanation:

The balanced chemical equation for the reaction between Na4XeO6 and Mn(NO3)2 6H2O can be written as:

Na4XeO6 + 2Mn(NO3)2 6H2O + 6NaOH -> 2MnO2 + 4NaNO3 + 4H2O + 4NaXeO3

To find the mass of Mn(NO3)2 6H2O that reacts with 125.0 mL of a 0.1717 M basic solution of Na4XeO6, we can use the formula:

Mass = (Volume x Concentration x Molar Mass) / 1000

Plugging in the values, we get:

Mass = (125.0 mL x 0.1717 M x 357.03 g/mol) / 1000 = 7.21 g

Therefore, the correct answer is d. 6.55 g