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3 years ago

Hydrogen can be prepared on a small scale by the hydrolysis of metal hydrides:

Chemistry

1 answer:

lara [203]3 years ago

6 0

Answer:

Ca²⁺ + 2H⁻ + 2(2H⁺ + O²⁻) ----> Ca²⁺ + 2O²⁻ + 2H⁺ + 4H⁰

(2 electrons lost) ----> (2 electrons gained)

Explanation:

The equation of the reaction is as follows: CaH₂ + 2H₂0 —> Ca(OH)₂ + 2H₂

The oxidation numbers of each of the atoms in the molecules are as follows:

In CaH₂, calcium has an oxidation number of +2, while hydrogen has an oxidation number of -1 ; Ca²⁺, 2H⁻

In 2H₂0, hydrogen has an oxidation number of +1 while oxygen has an oxidation number of -2 ; H⁺, O²⁻

In Ca(OH)₂, calcium has an oxidation number of +2 while oxygen has an oxidation number of -2 and hydrogen has an oxidation number of +1 ; Ca²⁺, O²⁻, 2H⁺

In 2H₂, each hydrogen atom has an oxidation number of 0 ; 4 H⁰

Ca²⁺ + 2H⁻ + 2(2H⁺ + O²⁻) ----> Ca²⁺ + 2O²⁻ + 2H⁺ + 4H⁰

On the left hand side, the hydride ion gives up two electrons to become neutral hydrogen atoms.

On the right hand, two hydrogen atoms accept two electrons to form neutral atoms.

These neutral hydrogen atoms pair up to form two hydrogen gas molecules.

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2 years ago

A student is studying the boiling points of six different liquids and finds results with consistently higher boiling points than

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2 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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3 years ago

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Likurg_2 [28]

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3 years ago

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KonstantinChe [14]

Answer:

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3 years ago