Consider the equation:
2 Na + 2 H₂O → 2 NaOH + H₂
If 92.0 g of sodium is reacted with 76.0 g of water until the reaction goes to completion,which reactant will remain and in what quantity?

Question

Quizplus · Accepted Answer

First, we need to balance the chemical equation to determine the mole ratio between Na and H2O:
2 Na + 2 H2O -> 2 NaOH + H2
Next, we need to determine which reactant will limit the amount of product produced (i.e. which reactant is the limiting reagent). To do this, we convert the mass of each reactant to moles using their respective molar masses:

92.0 g Na x (1 mol Na/22.99 g Na) = 4.00 mol Na
76.0 g H2O x (1 mol H2O/18.02 g H2O) = 4.22 mol H2O

According to the equation, it takes 2 moles of Na for every 2 moles of H2O. This means that 2.00 moles of Na would be required to completely react with 2.00 moles of H2O. Since we have 4.00 moles of Na, there is more than enough to completely react with the available H2O.

However, we only have 4.22 moles of H2O, which is less than the theoretical amount required for complete reaction. Therefore, H2O is the limiting reagent and the amount of Na remaining at the end of the reaction can be calculated as follows:

2.00 mol H2O x (2 mol Na/2 mol H2O) x (22.99 g Na/1 mol Na) = 45.98 g Na

Subtracting the amount of Na consumed from the initial amount of Na gives us the amount of Na that remains:
92.0 g Na - 45.98 g Na = 46.02 g Na

Therefore, the correct answer is (C) 3.9 g water, since all of the Na will be consumed in the reaction and some water will remain unreacted.

Consider the Equation: 2 Na + 2 H2O → 2 NaOH + H2

Consider the Equation: 2 Na + 2 H₂O → 2 NaOH + H₂