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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

How many grams of H2 can be produced by reaction of 1.80 g Al and 6.00 g H2SO4? 2 Al-example-1
User Dan Grahn
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4.9k points

1 Answer

3 votes

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

Step-by-step explanation:

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


\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} .....(1)

  • For aluminium:

Given mass of aluminium = 1.80 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:


\text{Moles of aluminium}=(1.80g)/(27g/mol)=0.066mol

  • For sulfuric acid:

Given mass of sulfuric acid = 6.00 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:


\text{Moles of sulfuric acid}=(6.00g)/(98g/mol)=0.061mol

The given chemical equation follows:


2Al+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2

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 =
(2)/(3)* 0.061=0.0408mol 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 =
(3)/(3)* 0.061=0.061moles 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:


0.061mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.061mol* 2g/mol)=0.122g

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

User Jimx
by
4.2k points