Upon decomposition of 90% of the initial mass of Me (OH) 3, water was obtained with a mass

| September 17, 2021 | education

With the decomposition of 90% of the initial mass of Me (OH) 3, water was obtained with a mass of 4.86 g and hydroxide with a mass of 2.8 g remained, install the metal.

Let us write the reaction equation for the decomposition of hydroxide:

2Me (OH) 3 = Me2O3 + 3H2O.

Let’s calculate the mass of the initial hydroxide:

m (Me (OH) 3) = m • W = 2.8 • 10 = 28 g.

This means that the mass of water with a complete 100% decomposition of the hydroxide will be:

m (H2O) = 4.86 • (100/90) = 5.4 g.

Mole amount of water:

n (H2O) = 5.4 / 18 = 0.3 mol.

And from the reaction equation it follows that the amount of hydroxide is 1.5 times less than that of water. Accordingly, the number of moles of Me (OH) 3 will be:

n (Me (OH) 3) = 0.3 / 1.5 = 0.2 mol.

Molar mass of hydroxide:

M (Me (OH) 3) = 28 / 0.2 = 140 g / mol.

Let’s calculate the molar mass of the metal:

M (Me) = 140 – (3 * 17) = 89 g / mol,

This metal is yttrium (Y).



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