What is the limiting reagent when 10.8 g water reacts with 4.7 g Na to produce NaOH and H2?

The first step to this problem is to write out reaction and balance it.

H2O + Na ---> NaOH + H2

The balance reaction looks like the following

2H2O +2 Na--->2NaOH + H2

This lets you know that the mole ratio between H2O and Na for this reaction to proceed must must be equal.

Convert grams of each to moles

10.8 g of H2O/18 g of H20/mole = number of moles H2O

4.7 g of Na/23.0 g of Na/mole = number of moles of Na

The compound with the least number of moles is the limiting reagent.

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Well, in this case, the limiting reagent is like the bouncer at a party. It decides who gets in and who gets left out. So, let's check out the guest list.

We have 10.8 grams of water and 4.7 grams of Na. To determine the limiting reagent, we need to compare the number of moles of each substance.

The molar mass of water (H2O) is approximately 18 g/mol (2 grams per mole for hydrogen and 16 grams per mole for oxygen).

So, for water, we have:
10.8 g / 18 g/mol ≈ 0.6 moles of water.

The molar mass of sodium (Na) is approximately 23 g/mol.

So, for sodium, we have:
4.7 g / 23 g/mol ≈ 0.2 moles of Na.

Now, let's look at the balanced chemical equation:
2Na + 2H2O -> 2NaOH + H2

According to the equation, we need 2 moles of Na for every 2 moles of water. So, the ratio is 1:1.

Since we have 0.6 moles of water and 0.2 moles of Na, we can see that we have more water than Na. In other words, the bouncer is telling the water to chill and let the Na take the lead.

Therefore, Na is the limiting reagent in this reaction. It's the star of the show, while water just has a supporting role.

To determine the limiting reagent in a chemical reaction, you need to compare the amount of reactants present and calculate how much product each reactant can produce. The reactant that produces less product is the limiting reagent.

Let's begin by writing the balanced chemical equation for the reaction:
2Na + 2H2O → 2NaOH + H2

To find the limiting reagent, you need to convert the given masses of water and sodium (Na) to moles using their molar masses. The molar masses are:
H2O = 18 g/mol
Na = 23 g/mol

For water (H2O):
10.8 g H2O × (1 mol H2O / 18 g H2O) ≈ 0.6 mol H2O

For sodium (Na):
4.7 g Na × (1 mol Na / 23 g Na) ≈ 0.2 mol Na

Next, you need to determine the mole ratio between the limiting reagent and the product(s) in the balanced equation. Looking at the coefficients in the equation, you can see that it takes 2 moles of Na to produce 1 mole of H2 and 2 moles of NaOH.

Now, let's calculate the number of moles of H2 and NaOH produced from each reactant:

From 0.6 mol H2O:
H2 produced = 0.6 mol H2O × (1 mol H2 / 2 mol H2O) = 0.3 mol H2
NaOH produced = 0.6 mol H2O × (2 mol NaOH / 2 mol H2O) = 0.6 mol NaOH

From 0.2 mol Na:
H2 produced = 0.2 mol Na × (1 mol H2 / 2 mol Na) = 0.1 mol H2
NaOH produced = 0.2 mol Na × (2 mol NaOH / 2 mol Na) = 0.2 mol NaOH

Comparing the moles of product produced, we can see that water (H2O) will produce a maximum of 0.3 moles of H2 and 0.6 moles of NaOH. On the other hand, sodium (Na) will only produce a maximum of 0.1 moles of H2 and 0.2 moles of NaOH.

Therefore, the limiting reagent is sodium (Na) because it produces less of the desired product (H2 and NaOH) compared to water (H2O).