How many grams of H2 can be produced by reaction of 1.80 g Al and 6.00 g
H2SO4?
2 Al + 3 H2SO4 => Al2(SO4)3 + 3 H2

0.0617
0.101
0.312
0.122
0.202

The amount of hydrogen gas that could be produced by the chemical reaction is 0.122 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of aluminium = 1.80 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

Given mass of sulfuric acid = 6.00 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of aluminium

So, 0.061 moles of sulfuric acid will react with = of aluminium

As, given amount of aluminium is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of sulfuric acid produces 3 moles of hydrogen gas

So, 0.061 moles of sulfuric acid will produce = of hydrogen gas

Now, calculating the mass of hydrogen gas from equation 1, we get:

Molar mass of hydrogen gas = 2 g/mol

Moles of hydrogen gas = 0.061 moles

Putting values in equation 1, we get:

Hence, the amount of hydrogen gas that could be produced by the chemical reaction is 0.122 grams

the dependant variable is spilled water

the independent is the heat