Question

8 fill in the blank 1 point 8. type your answer... n₂ + type your answer... o₂ → type your answer... n₂o₅

Explanation:

Step1: Balance nitrogen atoms

On the left side, we have \( N_2 \), and on the right side, we have \( N_2O_5 \) (2 N atoms). So N is balanced for now.

Step2: Balance oxygen atoms

Right side: \( N_2O_5 \) has 5 O atoms. Left side: \( O_2 \) has 2 O atoms. Let the coefficient of \( O_2 \) be \( x \) and \( N_2O_5 \) be \( y \). For O: \( 2x = 5y \). Let's find the least common multiple of 2 and 5, which is 10. So \( x=\frac{5}{2} \), \( y = 2 \)? Wait, no, we want whole number coefficients. Multiply the entire equation by 2 to eliminate fractions.

Step3: Adjust coefficients for whole numbers

Multiply each term by 2: \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \)? Wait, no, wait. Wait, original left: \( N_2 \) and \( O_2 \), right: \( N_2O_5 \). Let's balance step by step.

Let the equation be \( aN_2 + bO_2 ightarrow cN_2O_5 \).

For N: \( 2a = 2c \) (since left has 2a N, right has 2c N) ⇒ \( a = c \).

For O: \( 2b = 5c \). Let's take \( c = 2 \), then \( 2b = 10 \) ⇒ \( b = 5 \), and \( a = 2 \)? Wait, no, wait: if \( a = 1 \), then \( c = 1 \), so \( 2b = 5(1) \) ⇒ \( b = \frac{5}{2} \). But we need whole numbers, so multiply all coefficients by 2: \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \)? Wait, no, that would give 4 N on left and 4 N on right (2*2=4), and 10 O on left (5*2=10) and 10 O on right (2*5=10). Wait, but the problem's right side is \( N_2O_5 \), not \( 2N_2O_5 \). Wait, maybe I made a mistake. Wait, let's check again.

Wait, the product is \( N_2O_5 \) (1 molecule). So for N: left has \( N_2 \) (2 N), right has \( N_2O_5 \) (2 N). So N is balanced with coefficient 1 for \( N_2 \) and 1 for \( N_2O_5 \). Now O: right has 5 O, left has \( O_2 \) (2 O per molecule). So coefficient of \( O_2 \) should be \( \frac{5}{2} \). But we can't have fractions, so multiply the entire equation by 2 to get whole numbers: \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \). Wait, but the problem's blanks are for coefficients. Let's see the original blanks: first blank (coefficient of \( N_2 \)), second (coefficient of \( O_2 \)), third (coefficient of \( N_2O_5 \)).

Wait, let's do it properly. Let the equation be:

\( \text{[coefficient]} N_2 + \text{[coefficient]} O_2 ightarrow \text{[coefficient]} N_2O_5 \)

Balance N: left has 2*[N2 coefficient], right has 2*[N2O5 coefficient]. So [N2 coefficient] = [N2O5 coefficient].

Balance O: left has 2*[O2 coefficient], right has 5*[N2O5 coefficient]. So 2*[O2] = 5*[N2O5]. Let [N2O5] = 2 (to make O even), then [O2] = 5, [N2] = 2. Wait, but if [N2O5] = 1, then [O2] = 5/2, [N2] = 1. But we need whole numbers, so the balanced equation with whole numbers is \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \)? No, wait, no: wait, \( N_2 + \frac{5}{2}O_2 ightarrow N_2O_5 \). Multiply all by 2: \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \). But the problem's product is \( N_2O_5 \), not \( 2N_2O_5 \). Wait, maybe the problem allows fractional coefficients? No, usually we use whole numbers. Wait, maybe I messed up. Wait, let's check the number of atoms:

Left: \( N_2 \) (2 N) + \( O_2 \) (2 O per molecule). Right: \( N_2O_5 \) (2 N, 5 O).

So for O: 2*[O2] = 5*[N2O5]. Let [N2O5] = 2, then [O2] = 5, [N2] = 2. Then left: 2*2=4 N, 5*2=10 O. Right: 2*2=4 N, 2*5=10 O. That works. So the coefficients are 2, 5, 2? Wait, no, wait: \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \). But the problem's right side is \( N_2O_5 \), not \( 2N_2O_5 \). Wait, maybe the problem has a typo, or maybe I'm overcomplicating. Wait, let's do it with the product as \( N_2O_5 \) (1 molecule). Then:

N: 2 (left) = 2 (right) → coefficient of \( N_2 \) is 1.

O: 2*[O2] = 5*1 → [O2] = 5/2.

So the equation is \( 1N_2 + \frac{5}{2}O_2 ightarrow 1N_2O_5 \). But since we can't have fractions, we multiply all coefficients by 2 to get \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \). But the problem's blanks are for the coefficients in front of \( N_2 \), \( O_2 \), and \( N_2O_5 \). So:

First blank (coefficient of \( N_2 \)): 2? Wait, no, wait, maybe the problem expects the smallest whole number coefficients, but if we take the product as 2 \( N_2O_5 \), then:

Wait, let's check the original problem again. The reaction is \( N_2 + O_2 ightarrow N_2O_5 \). We need to balance it.

Balancing steps:

1. N: 2 on left, 2 on right (in \( N_2O_5 \)) → balanced for N with coefficient 1 for \( N_2 \) and 1 for \( N_2O_5 \).

2. O: 2 on left (in \( O_2 \)), 5 on right (in \( N_2O_5 \)) → to balance O, find LCM of 2 and 5, which is 10. So coefficient of \( O_2 \) is 5 (2*5=10 O), coefficient of \( N_2O_5 \) is 2 (5*2=10 O), and coefficient of \( N_2 \) is 2 (2*2=4 N, and \( N_2O_5 \) has 2*2=4 N). Wait, now N is balanced (4 N on left, 4 N on right), O is balanced (10 O on left, 10 O on right). So the balanced equation is \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \). But the problem's product is \( N_2O_5 \), not \( 2N_2O_5 \). Wait, maybe the problem has a mistake, or maybe I misread. Wait, no, the product is \( N_2O_5 \), so let's see:

If we have \( 1N_2 + \frac{5}{2}O_2 ightarrow 1N_2O_5 \), but we can't have fractions. So the correct balanced equation with whole numbers is \( 2N_2 + 5O_2 ightarrow 2N_2O_5 \). But the problem's blanks are for the coefficients in front of \( N_2 \), \( O_2 \), and \( N_2O_5 \). So:

First blank (N2 coefficient): 2

Second blank (O2 coefficient): 5

Third blank (N2O5 coefficient): 2

Wait, but let's verify:

Left: 2*N2 = 4 N, 5*O2 = 10 O

Right: 2*N2O5 = 4 N, 10 O

Yes, that's balanced. So the coefficients are 2, 5, 2.

Answer:

First blank: 2, Second blank: 5, Third blank: 2