Question

For the given reaction, if we react 18.4 grams of Al2S3 and 9.3 grams of H2O , which is the limiting reactant?Al2S3 + 6 H2O → 2Al(OH)3 + 3 H2SSelect one:a.Al2S3b.H2Sc.H2Od.Al(OH)3I understand the first part and have H20 = 0.51667 and Al2S3 = 0.122 moles but not sure what to do next

Answer 1

Answer:

$C\text{ : H}_2O$

Explanation:

Here, we want to get the limiting reactant

The limiting reactant is the reactant that produces less number of moles of the product

Firstly, we need to get the number of moles of each of the reactants

To get this, we divide the mass of the reactants by the molar mass of the reactants

The molar mass of Al2S3 is 150 g/mol

The number of moles is thus:

$\frac{18.4}{150}\text{ = 0.123 mol}$

Now, let us get the number of moles in the products

For Aluminum hydroxide we have 2 * 0.123 = 0.246 mol

For Hydrogen Sulfide, we have 3 * 0.123 = 0.369 mol

Now, let us get for water

We start by getting the number of moles. We divide the mass of water given by the molar mass of water which is 18 g/mol

That gives us:

$\frac{9.3}{18}\text{ = 0.517 mol}$

Now, from the equation of reaction:

6 moles of water gave 2 moles of Aluminum hydroxide

0.517 mol of water will give:

$\frac{0.517\text{ }\times2}{6}\text{ = 0.172 mol}$

For Hydrogen sulfide:

We have the number of moles as:

$\frac{0.517\text{ }\times\text{ 3}}{6}\text{ = 0.259 mol}$

Now, comparing the number of moles, the number of moles of products produced by water is smaller and that means water is the limiting reactant