Solve this stoichiometry problem 2 2 unread replies. 2 2 replies. solve the following problem showing all of your work and including all units. you can solve the problem in several steps or as a tandem conversion problem. how many grams of br2 are needed to completely convert 15.0 g al to albr3 ? al (s) + br2 (l) --> albr3 (hint: the equation is not balanced)

Answer : For, this question,

The balanced equation is; 2Al + 3 ----> 2Al

We can now, calculate how many grams of will be needed to completely convert 15 g of Al into Al.

Here, the reaction produces, 2 moles of Al

So, For 15.0 g of Al; we can calculate;

(15.0 g Al) X (1 mole Al / 26.98 g Al) X (3 moles / 2 moles Al) X (159.81 g / 1 mole )

= 133 g of will be needed.

Therefore, 133g of will be needed to completely convert 15 g of Al into Al.

Answer : The mass of needed are, 131.8 grams

Solution : Given,

Mass of Al = 15 g

Molar mass of Al = 27 g/mole

Molar mass of = 159.8 g/mole

First we have to calculate the moles of Al.

Now we have to calculate the moles of

The given balanced reaction will be,

From the balanced reaction, we conclude that

As, 2 moles of Al react with 3 moles of

So, 0.55 mole of Al react with of

Now we have to calculate the mass of

Therefore, the mass of needed are, 131.8 grams

Explanation:

A prussian blue colouration indicates the presence of nitrogen in the compound.

If nitrogen is present in the compound, the sodium extract would contain sodium cyanide formed during fusion.

On adding the required reagents, sodium cyanide reacts to form ferric-ferrocyanide which has prussian blue colour.