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Basile [38]
3 years ago
13

A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w

as precipitated as silver bromide, AgBr. The mass of the AgBr(s) was found to be 23.0052 grams. What is the identity of the unknown metal? Justify your answer with calculations.
Physics
1 answer:
barxatty [35]3 years ago
8 0

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

Moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}

Moles\ of\ AgBr= 0.1225\ mol

The reaction taking place is:

MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4

From the reaction,

4 moles of AgBr is produced when 1 mole of MBr_4 undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of MBr_4 undergoes reaction

0.1225 mole of AgBr is produced when \frac {1}{4}\times 0.1225 mole of MBr_4 undergoes reaction

Moles of MBr_4 got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of MBr_4 = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

0.030625 = \frac{12.5843}{319.616 + x}

Solve for x,

we get, x = 91.2999 g/mol

<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

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