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

The mass of aluminium required is 1.38 grams.

Explanation:

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

    .....(1)

Given mass of = 4.40 g

Molar mass of = 229 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

3 moles of reacts with 8 moles of aluminium

So, 0.0192 moles of will react with = 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:

Hence, the mass of aluminium required is 1.38 grams.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2