Question

1. use the following chemical equation to solve the 3 step problem and for completing the table. balance the equation. $ce{mg + n_{2} -> mg_{3}n_{2}}$

a. how many grams of product can be made from 25 g of mg and an excess of molecular nitrogen?
\ti. convert grams of mg to moles of mg using molar mass.
\tii. convert moles of mg to moles of product using coefficients.
\tiii. convert moles of product to grams of product.

b.

moles mg	moles $ce{n_{2}}$	moles $ce{mg_{3}n_{2}}$	excess mg	excess $ce{n_{2}}$	limiting reactant
6 moles	1 moles
6.6 moles		3.2 moles	0
		4 moles	1 mole	0
		3 moles	0	3 moles

Explanation:

First, balance the given chemical equation:

Step1: Balance Mg and N atoms

$$3\text{Mg} + \text{N}_2 ightarrow \text{Mg}_3\text{N}_2$$

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Part a:

Step1: Convert g Mg to mol Mg

Molar mass of Mg = $24.305\ \text{g/mol}$
$$\text{Moles of Mg} = \frac{25\ \text{g}}{24.305\ \text{g/mol}} \approx 1.029\ \text{mol}$$

Step2: Convert mol Mg to mol product

Mole ratio $\text{Mg}:\text{Mg}_3\text{N}_2 = 3:1$
$$\text{Moles of Mg}_3\text{N}_2 = 1.029\ \text{mol} \times \frac{1}{3} \approx 0.343\ \text{mol}$$

Step3: Convert mol product to g product

Molar mass of $\text{Mg}_3\text{N}_2 = (3 \times 24.305) + (2 \times 14.007) = 100.929\ \text{g/mol}$
$$\text{Mass of Mg}_3\text{N}_2 = 0.343\ \text{mol} \times 100.929\ \text{g/mol} \approx 34.6\ \text{g}$$

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Part b:

Use the balanced equation $3\text{Mg} + \text{N}_2
ightarrow \text{Mg}_3\text{N}_2$ for all table calculations:

Row1 (3 mol Mg, 6 mol N₂):

Step1: Find moles of product

Mg is limiting; $\text{Moles of Mg}_3\text{N}_2 = \frac{3}{3} = 1\ \text{mol}$

Step2: Find excess reactant

Used $\text{N}_2 = 1\ \text{mol}$, Excess $\text{N}_2 = 6-1=5\ \text{mol}$
Excess Mg = 0; Limiting = Mg

Row2 (6 mol Mg, 1 mol N₂):

Step1: Find moles of product

N₂ is limiting; $\text{Moles of Mg}_3\text{N}_2 = 1\ \text{mol}$

Step2: Find excess reactant

Used Mg = $3 \times 1=3\ \text{mol}$, Excess Mg = 6-3=3\ \text{mol}$
Excess N₂ = 0; Limiting = N₂

Row3 (6.6 mol Mg, 3.2 mol Mg₃N₂):

Step1: Find moles of N₂

Used $\text{N}_2 = 3.2\ \text{mol}$, Excess $\text{N}_2 = (\text{Total N}_2) - 3.2$; Used Mg = $3 \times 3.2=9.6\ \text{mol}$, but given Mg=6.6 mol (limiting), so $\text{Total N}_2 = 3.2 + \text{Excess N}_2$. Since Mg is limiting, $\text{Total N}_2 = 3.2 + \left(\frac{6.6}{3} - 3.2
ight)^{-1}$ corrected: $\text{Total N}_2 = 3.2 + \text{Excess N}_2$, Used N₂ = $\frac{6.6}{3}=2.2\ \text{mol}$, so $\text{Total N}_2 = 2.2 + \text{Excess N}_2$; given product=3.2 is inconsistent, use Mg limiting: $\text{Actual product} = \frac{6.6}{3}=2.2\ \text{mol}$, but table says 3.2, so $\text{Total N}_2 = 3.2\ \text{mol}$, Excess $\text{N}_2=3.2-2.2=1.0\ \text{mol}$; Limiting = Mg

Row4 (1 mol excess Mg, 4 mol Mg₃N₂):

Step1: Find total moles of Mg

Used Mg = $3 \times 4=12\ \text{mol}$, Total Mg = $12+1=13\ \text{mol}$

Step2: Find moles of N₂

Used $\text{N}_2 = 4\ \text{mol}$, Excess N₂=0, so Total N₂=4\ \text{mol}$; Limiting = N₂

Row5 (0 excess Mg, 3 mol Mg₃N₂, 3 mol excess N₂):

Step1: Find total moles of Mg

Used Mg = $3 \times 3=9\ \text{mol}$, Total Mg=9\ \text{mol}$

Step2: Find total moles of N₂

Used $\text{N}_2=3\ \text{mol}$, Total N₂=3+3=6\ \text{mol}$; Limiting = Mg

Answer:

Balanced equation: $3\text{Mg} + \text{N}_2
ightarrow \text{Mg}_3\text{N}_2$
a. $\approx 35\ \text{g}$ (rounded to 2 significant figures)
b. Completed table:

Moles Mg	Moles N₂	Moles Mg₃N₂	Excess Mg	Excess N₂	Limiting Reactant
6 moles	1 mole	1 mole	3 moles	0	N₂
6.6 moles	3.2 moles	2.2 moles*	0	1.0 mole	Mg
13 moles	4 moles	4 moles	1 mole	0	N₂
9 moles	6 moles	3 moles	0	3 moles	Mg

*Note: The given 3.2 moles of $\text{Mg}_3\text{N}_2$ is inconsistent with 6.6 moles of Mg; the theoretical maximum product from 6.6 moles of Mg is 2.2 moles.