Question

For every liter of sea water that evaporates, 3.7 g of magnesium hydroxide are produced. How many liters of sea water must evaporate to produce 5.00 moles of magnesium hydroxide?

Answer 1

We can use Algebra and ratios to find this value, but first we must convert 3.7g to moles.

First we find the molar mass of Mg(OH)₂
Mg + 2O + 2H = 24.31 + 2×16 + 2×1.01 = 58.33 gmol⁻¹.

Then find the number of moles: n = m/M = 3.7 / 58.33 = 0.0634 mol

So now we know that 0.0634 mol are produced per litre of sea water.
Let the amount of water in litres needed to evapourate to produce 5.00 moles be x.

We can express this as a ratio.

0.0634 mol / 1 l = 5 mol / x l.
Solve for x by cross-multiplication:
0.0634x = 5×1
x = 5/0.0634 = 78.8 litres. (rounded to three significant figures)

Hope this helps!

Answer 2

The liters of sea water must evaporate to produce 5.00 moles of magnesium hydroxide - 78.8 litres.

A mole is a special unit of quantity of a chemical species that contains exactly Avogadro's number of particles.

• The mass of one mole of an element or a compound, called the molar mass, can be found from a periodic table.
• The unit of molar mass is g/mol.

=> the molar mass of Mg(OH)₂

Mg + 2O + 2H

= (24.31) + (2×16) + (2×1.01)

= 58.33 gmol⁻¹.

then,

=> the number of moles:

n = m/M

= $\frac{3.7}{58.33}$

= 0.0634 mol

=> the amount of water in liters needed to evapourate to produce 5.00 moles:

=> $\frac{0.0634}{1L} = \frac{5}{ x L}$

=> 0.0634x = 5×1

=> $x = \frac{5}{0.0634}$

=> x = 78.8 litres.

Thus, the liters of sea water must evaporate to produce 5.00 moles of magnesium hydroxide - 78.8 litres.

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