Question

How many g of Aluminum are needed to react completely with 4.40 gram of Mn304, according to the chemical equation below? 3 Mn304 8A>9 Mn 4 A1203 2.65 g 19.2 g 24.3 g ④ 1.54 g 1.38 g (6) 0.860 g ⑦ 3.91 g 8.70g

Answer 1

Answer: The mass of aluminium required is 1.38 grams.

Explanation:

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

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

Given mass of $Mn_3O_4$ = 4.40 g

Molar mass of $Mn_3O_4$ = 229 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Mn_3O_4=\frac{4.40g}{228.8g/mol}=0.0192mol$

For the given chemical equation:

$3Mn_3O_4+8Al\rightarrow 9Mn+4Al_2O_3$

By Stoichiometry of the reaction:

3 moles of $Mn_3O_4$ reacts with 8 moles of aluminium

So, 0.0192 moles of $Mn_3O_4$ will react with = $\frac{8}{3}\times 0.0192=0.0512mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1, we get:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.0512 moles

Putting values in equation 1, we get:

$0.0512mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.0512mol\times 27g/mol)=1.38g$

Hence, the mass of aluminium required is 1.38 grams.