Determine the name and formula of a hydrate with the following composition; 32.30%

Chemistry

1 answer:

Maslowich2 years ago

7 0

1. The formula of the hydrate is FePO₄•4H₂O

2. The name of the hydrate is Iron(iii) phosphate tetrahydrate.

1. Determination of the formula of the hydrate

H₂O = 32.3%
FePO₄ = 67.7%
Formula =?

The formula of the hydrate can be obtained as follow:

Divide by their molar mass

H₂O = 32.3 / 18 = 1.794

FePO₄ = 67.7 / 151 = 0.448

Divide by the smallest

H₂O = 1.794 / 0.448 = 4

FePO₄ = 0.448 / 0.448 = 1

Therefore, the formula of the hydrate is FePO₄•4H₂O

The name of the hydrate is Iron(iii) phosphate tetrahydrate

Learn more about empirical formula: brainly.com/question/16143068

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The burning of magnesium is a highly exothermic reaction. How many kilojoules of heat are released when 0. 75 mol of Mg burn in

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Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 0.75 moles of Mg are burned is 451.5 kJ.

<h3>Enthalpy of a chemical reaction</h3>

The enthalpy of a chemical reaction is known as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

<h3>Heat released in this case</h3>

In this case, the balanced reaction is:

2 Mg(s) + O₂ (g) → 2 MgO(s) + 1204 kJ

This equation indicates that when 2 moles of Mg reacts with 1 mole of O₂, 1204 kJ of heat is released.

When 0.75 moles of Mg are burned, then you can apply the following rule of three: if 2 moles of Mg releases 1204 kJ of heat, 0.75 moles of Mg releases how much heat?

$heat=\frac{0.75 moles of Mgx1204 kJ}{2 moles of Mg}$