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cupoosta [38]
3 years ago
15

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
______________________
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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An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the
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Approximately 81.84\%.

Explanation:

Balanced equation for this reaction:

{\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm  NaCl}\, (aq) + {\rm CaCO_{3}}\, (s).

Look up the relative atomic mass of elements in the limiting reactant, \rm CaCl_{2}, as well as those in the product of interest, \rm CaCO_{3}:

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Calculate the formula mass for both the limiting reactant and the product of interest:

\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}.

\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the quantity of the limiting reactant (\rm CaCl_{2}) available to this reaction:

\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}.

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant (\rm CaCl_{2}) and the product ({\rm CaCO_{3}}) are both 1. Thus:

\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1.

In other words, for every 1\; \rm mol of \rm CaCl_{2} formula units that are consumed, 1\; \rm mol\! of \rm CaCO_{3} formula units would (in theory) be produced. Thus, calculate the theoretical yield of \rm CaCO_{3}\! in this experiment:

\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}.

Calculate the theoretical yield of this experiment in terms of the mass of \rm CaCO_{3} expected to be produced:

\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}.

Given that the actual yield in this question (in terms of the mass of \rm CaCO_{3}) is 13.19\; \rm g, calculate the percentage yield of this experiment:

\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}.

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