if 25.2 g of aluminum (AW=27.0) are mixed with 125 g of sulfuric acid(MM=98.1)then how many grams of aluminum sulfate will be produced?

2Al+ 3 H2SO4 --> 3H2 + Al2(SO4)3

convert aluminum to moles, and sulfuric acid to moles.

According to the balanced equation, you need 1.5 :1 moles Sulfuricacid:aluminum
If you have a ratio greater than this,aluminum is your limiting reageant.
If less, sulfuric acid is limiting.

I haven't worked those ratios, so do it.
Take the moles of your limiting reageant, if it was aluminum, you should get 1/2 those moles of Aluminum sulfate. Convert to grams.
If the limiting reageant was acid, you should get 2/3 of the moles of sulfuric acid as the moles of aluminum sulfate. Convert to grams.

To determine the amount of aluminum sulfate produced, we need to determine the limiting reactant. This is the reactant that is completely consumed in the reaction, determining the maximum amount of product that can be formed.

First, we need to calculate the number of moles of aluminum and sulfuric acid:

Moles of Aluminum = Mass of Aluminum / Atomic Weight of Aluminum
Moles of Aluminum = 25.2 g / 27.0 g/mol
Moles of Aluminum = 0.933 mol

Moles of Sulfuric Acid = Mass of Sulfuric Acid / Molar Mass of Sulfuric Acid
Moles of Sulfuric Acid = 125 g / 98.1 g/mol
Moles of Sulfuric Acid = 1.274 mol

Next, we need to determine the stoichiometric ratio between aluminum and aluminum sulfate. From the balanced equation, we can see that 2 moles of aluminum react to produce 1 mole of aluminum sulfate. Therefore, the moles of aluminum sulfate produced can be calculated as follows:

Moles of Aluminum Sulfate = (Moles of Aluminum) / 2
Moles of Aluminum Sulfate = 0.933 mol / 2
Moles of Aluminum Sulfate = 0.467 mol

Finally, we can calculate the mass of aluminum sulfate produced using the moles and molar mass:

Mass of Aluminum Sulfate = (Moles of Aluminum Sulfate) * (Molar Mass of Aluminum Sulfate)
Mass of Aluminum Sulfate = 0.467 mol * (27.0 g/mol + 3*(32.1 g/mol + 16.0 g/mol))
Mass of Aluminum Sulfate = 0.467 mol * 342.2 g/mol
Mass of Aluminum Sulfate = 159.7 g

Therefore, approximately 159.7 grams of aluminum sulfate will be produced.

To find the mass of aluminum sulfate produced, we need to calculate the limiting reactant first. The limiting reactant is the one that will be completely consumed, stopping the reaction once it is used up. The reactant that is not limiting is called the excess reactant.

To determine the limiting reactant, we need to compare the number of moles of aluminum and sulfuric acid used in the reaction.

First, let's calculate the number of moles of aluminum and sulfuric acid:

Number of moles of aluminum = mass of aluminum / molar mass of aluminum
Number of moles of aluminum = 25.2 g / 27.0 g/mol

Number of moles of sulfuric acid = mass of sulfuric acid / molar mass of sulfuric acid
Number of moles of sulfuric acid = 125 g / 98.1 g/mol

Next, we can use the balanced equation to determine the ratio of moles between aluminum and sulfuric acid. From the balanced equation:
2Al + 3H2SO4 -> 3H2 + Al2(SO4)3

The ratio of moles between aluminum and sulfuric acid is 2:3.

Now, let's calculate the moles of aluminum sulfate that can be produced if all the aluminum reacts:

Moles of aluminum sulfate = (moles of aluminum) × (moles of aluminum sulfate produced per mole of aluminum)
Moles of aluminum sulfate = (moles of aluminum) × (1 mole of aluminum sulfate / 2 moles of aluminum)

Finally, we can calculate the mass of aluminum sulfate produced by multiplying the moles of aluminum sulfate by its molar mass:

Mass of aluminum sulfate = (moles of aluminum sulfate) × (molar mass of aluminum sulfate)

Hope this explanation helps in solving the problem!