An excess of sodium hydroxide is treated with 2.3 L of dry hydrogen iodide gas measured at STP to produce sodium iodide and water.What is the mass of sodium iodide formed in the process?
A) 1.15 g
B) 30.8 g
C) 15.4 g
D) 19 g
E) 345 g

Question

Quizplus · Accepted Answer

We can use the balanced chemical equation to determine the limiting reactant and the amount of sodium iodide formed:

2NaOH + 2HI -> 2NaI + 2H2O

From the equation, we see that 2 moles of sodium hydroxide react with 2 moles of hydrogen iodide to produce 2 moles of sodium iodide. Therefore, the mole ratio of NaOH to HI is 1:1.

To determine the limiting reactant, we need to first convert the volume of hydrogen iodide gas to moles. At STP, one mole of gas occupies 22.4 L. Therefore,

2.3 L H2I * (1 mol H2I / 22.4 L) = 0.103 moles H2I

Next, we need to calculate the number of moles of NaOH that reacts with the hydrogen iodide. The volume of NaOH is not given, but since it is in excess, we can assume that all of the hydrogen iodide reacts with the NaOH. Therefore,

0.103 moles H2I * (1 mol NaOH / 1 mol H2I) = 0.103 moles NaOH

Since the mole ratio of NaOH to NaI is 1:1, the number of moles of NaI formed is also 0.103. We can now use the molar mass of NaI to convert from moles to grams:

0.103 moles NaI * (149.89 g NaI / 1 mol NaI) = 15.4 g NaI

Therefore, the answer is choice C, 15.4 g of NaI is formed.

An Excess of Sodium Hydroxide Is Treated with 2