I\'m not sure if someone can answer this, because it\'s from a lab experiement.

Question

I'm not sure if someone can answer this, because it's from a lab experiement.

Ok, First, I'll explain what is going on with this question, it may be kinda long, but I want to try and provide as much information as I possibly can.

I have to set up a reaction where napthaelene and 2-bromo-2-methylpropane are combined in equimolar (1:1) stoichiometry.

First, I had to dispense 5.00 mL of 2-bromo-2-methylpropane. Next, I have to calculate the mass of napthaelene required to react stoichiometrically with the moles of 2-bromo-2-methylpropane that was placed in the flask (which was the 5.00 mL)

Then, I have to dispense that quantity, but I need to know what quantity to dispense, and that's where I'm stuck/confused..

Explanation / Answer

Solution :-

Using the volume and density of the 2-bromo-2-methylpropane lets calculate its mass

Volume = 5 ml

Density = 1.22 g/ml

Mass = volume * density

Mass of 2-bromo-2-methylpropane = 5.0 ml * 1.22 g per ml = 6.1 g

Now lets calculate the moles of 2-bromo-2-methylpropane

Moles= mass / molar mass

Moles of 2-bromo-2-methylpropane = 6.1 g / 137.02 g per mol

                                                                  = 0.04452 mol

Now as we know the mole ratio of the 2-bromo-2-methylpropane and naphthalene is 1 : 1

Therefore moles of naphthalene required for the reaction are same as moles of 2-bromo-2-methylpropane

Therefore the moles of naphthalene needed = 0.04452 mol

Now lets calculate the mass of the naphthalene needed

Mass = moles * molar mass

Mass of naphthalene = 0.04452 mol * 128.1705 g per mol

                                      = 5.71 g naphthalene

So the mass of the naphthalene needed = 5.71 g

So need to despense 5.71 g naphthalene

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