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nordsb [41]
3 years ago
8

If 75 grams of oxygen react, how many grams of aluminum are required?

Chemistry
1 answer:
german3 years ago
5 0

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂     →   2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

                          O₂         :          Al

                           3          :             4

                        2.34        :         4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

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Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.
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For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

                                ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where: 
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol

m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg


For the problem, 
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius
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3 years ago
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Answer:

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