Question

Magnesium nitrate, Mg(NO3)2, will decompose when heated to give a white solid and a mixture of gases. One of the gases released is an oxide of nitrogen, X. 7.4 g of anhydrous magnesium nitrate is heated until no further reaction takes place. What mass of X is produced? a 1.5 g b. 2.3 g c. 3.0 g d. 4.6 g

Answer 1

Answer:Hi!!!

X = NO2

Answer will be = 4.6

Explanation:Hope you get the answer..

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Answer 2

Answer:

The mass of gas X, i.e., $NO_{2}$ produced is $4.6g$.

Therefore, option d) $4.6g$ is correct.

Explanation:

Data given,

The reaction: $2Mg(NO_{3})_{2} \overset{\Delta}{\rightarrow} 2 MgO + 4NO_{2} + O_{2}$

The oxide gas, X, released is $NO_{2}$.

The mass of $Mg(NO_{3})_{2}$ = $7.4g$

The mass of gas X, i.e., $NO_{2}$ =?

From the above reaction, we can see that the $2 moles$ of $Mg(NO_{3})_{2}$ yield $4 moles$ of $NO_{2}$.

Therefore,

• $1mole$ of $Mg(NO_{3})_{2}$ yield = $2moles$ of $NO_{2}$

Now, the molar mass of $Mg(NO_{3})_{2}$ = $148g/mol$

And the molar mass of $NO_{2}$ = $46g/mol$

Also, the molar mass of $2$$NO_{2}$ = $2\times 46$ = $92g$

Therefore,

• $148g$ of $Mg(NO_{3})_{2}$ yield = $92g$ of $NO_{2}$
• $1g$ of $Mg(NO_{3})_{2}$ yield = $\frac{92}{148}g$ of $NO_{2}$

Also,

• $7.4g$ of $Mg(NO_{3})_{2}$ yield = $7.4\times \frac{92}{148}g$ of $NO_{2}$  = $4.6g$ of $NO_{2}$.

Hence, the mass of gas X, i.e., $NO_{2}$ produced = $4.6g$.