Question

For the following reaction, 101 grams of magnesium nitride are allowed to react with 144 grams of water. Mg3N2 (5) + 6 H20 (1) — 3 Mg(OH)2 (aq) + 2 NH2 (aq) What is the FORMULA for the limiting reagent? What is the maximum amount of magnesium hydroxide that can be formed? grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining

Answer 1

Answer : The formula of limiting reagent is,
`Mg_3N_2`.

The mass of
`Mg(OH)_2` is, 174 grams.

The mass of excess reactant is, 36 grams.

Solution : Given,

Mass of
`Mg_3N_2` = 101 g

Mass of
`H_2O` = 144 g

Molar mass of
`Mg_3N_2` = 101 g/mole

Molar mass of
`H_2O` = 18 g/mole

Molar mass of
`Mg(OH)_2` = 58 g/mole

First we have to calculate the moles of
`Mg_3N_2` and
`H_2O`.

`\text{ Moles of }Mg_3N_2=\frac{\text{ Mass of }Mg_3N_2}{\text{ Molar mass of }Mg_3N_2}=(101g)/(101g/mole)=1moles`

`\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=(144g)/(18g/mole)=8moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`Mg_3N_2+6H_2O\rightarrow 3Mg(OH)_2+2NH_3`

From the balanced reaction we conclude that

As, 1 mole of
`Mg_3N_2` react with 6 mole of
`H_2O`

So, given 1 moles of
`Mg_3N_2` react with 6 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
`Mg_3N_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`Mg(OH)_2`

From the reaction, we conclude that

As, 1 mole of
`Mg_3N_2` react to give 3 mole of
`Mg(OH)_2`

So, given 1 mole of
`Mg_3N_2` react to give 3 moles of
`Mg(OH)_2`

Now we have to calculate the mass of
`Mg(OH)_2`

`\text{ Mass of }Mg(OH)_2=\text{ Moles of }Mg(OH)_2* \text{ Molar mass of }Mg(OH)_2`

`\text{ Mass of }Mg(OH)_2=(3moles)* (58g/mole)=174g`

The mass of
`Mg(OH)_2` is, 174 grams.

Now we have to calculate the moles of excess reactant
`(H_2O)`.

Moles of excess reactant = 8 - 6 = 2 moles

Now we have to calculate the mass of excess reactant.

`\text{ Mass of }H_2O=\text{ Moles of }H_2O* \text{ Molar mass of }H_2O`

`\text{ Mass of }H_2O=(2moles)* (18g/mole)=36g`

The mass of excess reactant is, 36 grams.