Question

the sentence below is a word equation that describes a chemical reaction. write the balanced chemical equation for this reaction.
iron reacts with nitrogen to form iron(iii) nitride.
step 1
write the skeleton equation. begin this step by determining the basic structure of the reaction - that is, the number of reactants and products. tap below to toggle through and select one of the four options.
reactant #1 → product #1
check answer

Explanation:

Step1: Identify reactants and products

Reactants: Iron (Fe) and Nitrogen ($N_2$, since nitrogen is diatomic in its elemental form). Product: Iron(III) nitride. The formula for iron(III) nitride: Iron has a +3 charge, nitrogen has a -3 charge (in nitride), so the formula is $FeN$. Wait, no, let's correct that. The formula for iron(III) nitride: Fe has +3, N has -3? Wait, no, nitride is $N^{3 - }$, so to balance charges, Fe (III) is $Fe^{3+}$, so the formula is $FeN$? Wait, no, actually, the correct formula for iron(III) nitride is $Fe_3N_2$? Wait, no, let's do the criss - cross method. Fe³⁺ and N³⁻ (nitride ion). So the formula would be $FeN$? Wait, no, that can't be. Wait, nitrogen in elemental form is $N_2$, and when forming iron(III) nitride, let's re - evaluate.

Wait, the reaction is Iron (Fe) reacts with Nitrogen ($N_2$) to form Iron(III) nitride. Let's first write the skeleton equation. Reactants: Fe and $N_2$. Product: $FeN$? No, wait, the correct formula for iron(III) nitride is $Fe_3N_2$? Wait, no, let's check the oxidation states. Iron(III) is +3, nitrogen in nitride is -3. So the formula should be $FeN$ (since 3 and 3 cancel out). But when we balance the equation, let's start over.

Skeleton equation: $Fe + N_2
ightarrow FeN$

Now, balance the equation. Let's count the number of N atoms. On the left, we have 2 N atoms (from $N_2$), on the right, we have 1 N atom (from $FeN$). So we need to put a coefficient of 2 in front of $FeN$ to balance N: $Fe + N_2
ightarrow 2FeN$

Now, balance Fe. On the right, we have 2 Fe atoms (from 2$FeN$), so we put a coefficient of 2 in front of Fe on the left: $2Fe + N_2
ightarrow 2FeN$

Wait, but let's check the formula of iron(III) nitride again. Maybe I made a mistake. The correct formula for iron(III) nitride is $Fe_3N_2$. Let's re - do it. If the product is $Fe_3N_2$, then:

Skeleton equation: $Fe + N_2
ightarrow Fe_3N_2$

Balance N: On the left, 2 N; on the right, 2 N (in $Fe_3N_2$). Now balance Fe: On the right, 3 Fe, so put 3 in front of Fe on the left: $3Fe + N_2
ightarrow Fe_3N_2$

Ah, that's correct. Because iron(III) has a +3 charge, and nitride is $N^{3 - }$? Wait, no, the nitride ion is $N^{3 - }$, but in $Fe_3N_2$, let's check the charges. Fe is +3, so 3 Fe atoms give 3*(+3)=+9. N in $Fe_3N_2$: 2 N atoms, so 2*(-x)= - 9? No, that can't be. Wait, I think I messed up the formula of iron(III) nitride. The correct formula for iron(III) nitride is $FeN$? No, let's refer to the correct formula. Iron(III) nitride has the formula $FeN$? Wait, no, a quick check: Iron(III) nitride: Fe³⁺ and N³⁻ (nitride), so the formula is $FeN$ (1:1 ratio). But when we balance the reaction between Fe and $N_2$ to form $FeN$:

$Fe + N_2
ightarrow FeN$

Balance N: 2 on left, 1 on right. So $Fe + N_2
ightarrow 2FeN$

Balance Fe: 2 on right, so 2 on left: $2Fe + N_2
ightarrow 2FeN$

Now, check charges: Fe is +3, N is -3, so $FeN$ is neutral. And the reaction is between elemental Fe (oxidation state 0) and elemental $N_2$ (oxidation state 0) to form $FeN$ (Fe +3, N - 3). So the oxidation states change, which is a redox reaction, but the balancing is as above.

Wait, but maybe the correct formula is $Fe_3N_2$. Let's calculate the oxidation states in $Fe_3N_2$. Let the oxidation state of Fe be x. Then 3x + 2*(-3)=0 (since nitride is -3). 3x - 6 = 0 → 3x=6 → x = +2. But that's iron(II) nitride, not iron(III). So my initial mistake was in the formula of iron(III) nitride. The correct formula for iron(III) nitride is $FeN$ (Fe +3, N - 3) or $Fe_3N$? Wait, no, let's use the criss - cross method properly. Fe³⁺ and N³⁻: the charges are both 3, so the formula is $FeN$ (1:1). So the reaction is:

Reactants: Fe (s) and $N_2$ (g)

Product: $FeN$ (s)

Skeleton equation: $Fe + N_2
ightarrow FeN$

Now, balance N: There are 2 N atoms on the left, 1 on the right. So we put a coefficient of 2 in front of $FeN$:

$Fe + N_2
ightarrow 2FeN$

Now, balance Fe: There are 2 Fe atoms on the right, so we put a coefficient of 2 in front of Fe:

$2Fe + N_2
ightarrow 2FeN$

Let's check the number of each atom:

- Fe: Left: 2, Right: 2 (from 2$FeN$)
- N: Left: 2 (from $N_2$), Right: 2 (from 2$FeN$)

So the balanced chemical equation is $2Fe + N_2 = 2FeN$

Step2: Verify the balancing

Count the number of Fe atoms: left side has 2 (from 2Fe), right side has 2 (from 2FeN). Count the number of N atoms: left side has 2 (from $N_2$), right side has 2 (from 2FeN). So the equation is balanced.

Answer:

The balanced chemical equation is $\boldsymbol{2Fe + N_2 = 2FeN}$