To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.

We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'

$\frac{40 g Mg}{1} x\frac{1 mole Mg}{24.305 g Mg}$

In this, we can see that the unit 'grams' will cancel out to leave us with moles.

In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.

3 0

3 years ago

Find the volume of 0.100M hydrochloric acid necessary to react completely with 1.51g Al(OH)3.

shtirl [24]

Reaction equation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
Moles of Al(OH)₃:
moles = mass/Mr
= 1.51 / (27 + 17 x 3)
= 0.019
Molar ratio Al(OH)₃ : HCl = 1 : 3
Moles of HCl required = 0.019 x 3
=0.057
concentration = moles/volume
volume = 0.057 / 0.1
= 0.57 dm³
= 570 ml

3 0

3 years ago