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3 years ago

hat is the product when magnesium reacts with nitrogen? Mg(s) + N2(g) → Mg2N3(s) Mg3N(s) Mg3N2(s) MgN3(s)

Chemistry

2 answers:

Ipatiy [6.2K]3 years ago

5 0

3Mg + N₂ = Mg₃N₂

Mg₃N₂
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Pani-rosa [81]3 years ago

5 0

<u>Answer:</u> The product of the given reaction is $Mg_3N_2(s)$

<u>Explanation:</u>

When magnesium solid reacts with nitrogen gas, it leads to the formation of an ionic solid known as magnesium nitride.

The chemical equation for the reaction of magnesium and nitrogen gas follows:

$3Mg(s)+N_2(g)\rightarrow Mg_3N_2(s)$

By Stoichiometry of the reaction:

3 moles of magnesium solid reacts with 1 mole of nitrogen gas to produce 1 mole of magnesium nitride.

The given reaction is considered as synthesis reaction.

Hence, the product of the given reaction is $Mg_3N_2(s)$

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A separatory funnel contains the two immiscible liquids water and toluene. Use the given densities to determine which layer is o

damaskus [11]

Answer:

Top layer Toluene

Bottom layer Water

Explanation:

When two non-miscible liquids are put together, the one with the higher density will be on the bottom, while the one with the lower density will be on top.

Meaning that in this problem's case toluene would be on the top layer and water in the bottom layer.

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3 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

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3 years ago

What are some physical properties of metal

bixtya [17]

Physical properties of metal include shiny, ductile, opaque, malleable and good conduction of heat and electricity

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4 years ago

Can someone tell me how to draw a atomic model for Cd

tensa zangetsu [6.8K]

Explanation:

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3 years ago

The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th

horrorfan [7]

Answer:

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

Explanation:

$ClO ( g ) + O_3 ( g )\rightarrow Cl ( g ) + 2 O_2 ( g ),\Delta H^o_{1,rxn} =-122.8 kJ$ ..[1]

$2 O_3 ( g )\rightarrow 3O_2 ( g ),\Delta H^o_{2,rxn} = -285.3 kJ$ ..[2]

$O_3(g) + Cl(g)\rightarrow ClO (g)+O_2(g),\Delta H^o_{3,rxn}=?$ ..[3]

The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:

[2] - [1] = [3]

$\Delta H^o_{3,rxn}=\Delta H^o_{2,rxn}-\Delta H^o_{1,rxn}$

$=-285.3 kJ-(-122.8 kJ)=162.5 kJ$

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

8 0

3 years ago