Question

click here to access the interactive simulation. you will practice balancing chemical equations by playing an interactive game. after accessing the website, select the game box and start with level 1, followed by levels 2 and 3. the goal of these games is to have the same number of atoms on each side of the equation.

in level 1, you balanced the equation $\ce{n_{2} + h_{2} -> \boxed{}nh_{3}}$. what number did you place in front of ammonia ($\ce{nh_{3}}$)?

in level 2, you balanced the equation $\ce{c + h_{2}o -> ch_{4} + \boxed{}co_{2}}$. what number did you place in front of carbon dioxide ($\ce{co_{2}}$)?

in level 3, you balanced the equation $\ce{nh_{3} + o_{2} -> no_{2} + \boxed{}h_{2}o}$. what number did you place in front of water?

Explanation:

Level 1: Balancing \( \boldsymbol{\ce{N2 + H2 -> NH3}} \)

Step1: Balance Nitrogen atoms

On the left, we have \( \ce{N2} \) (2 N atoms). On the right, \( \ce{NH3} \) has 1 N atom. So we need 2 \( \ce{NH3} \) to balance N: \( \ce{N2 + H2 -> 2NH3} \).

Step2: Balance Hydrogen atoms

Now, right side has \( 2\times3 = 6 \) H atoms. Left side has \( \ce{H2} \), so we need 3 \( \ce{H2} \): \( \ce{N2 + 3H2 -> 2NH3} \). The coefficient for \( \ce{NH3} \) is 2.

Step1: Balance Carbon atoms

Left: 1 C (from \( \ce{C} \)). Right: 1 C (from \( \ce{CH4} \)) + 1 C (from \( \ce{CO2} \)) = 2 C. So we need 2 \( \ce{C} \) on left: \( \ce{2C + H2O -> CH4 + CO2} \).

Step2: Balance Hydrogen atoms

Right: \( \ce{CH4} \) has 4 H. Left: \( \ce{H2O} \) has 2 H per molecule. Let coefficient of \( \ce{H2O} \) be \( x \), so \( 2x = 4 \) → \( x = 2 \)? Wait, no. Wait, after C balance: \( \ce{2C + H2O -> CH4 + CO2} \). Now H: right has 4, left has \( 2y \) (from \( y \) \( \ce{H2O} \)). Wait, let's re - do. Let's balance H first. \( \ce{CH4} \) has 4 H, so \( \ce{H2O} \) needs 2 molecules (4 H). So \( \ce{C + 2H2O -> CH4 + CO2} \). Now C: left 1, right 1 (CH4) + 1 (CO2) = 2. So left C: 2. So \( \ce{2C + 2H2O -> CH4 + CO2} \)? No, wait. Wait, correct approach: Let coefficient of \( \ce{CO2} \) be \( a \), \( \ce{CH4} \) be \( b \), \( \ce{C} \) be \( c \), \( \ce{H2O} \) be \( d \). C: \( c = b + a \). H: \( 2d = 4b \). O: \( d = 2a \). Let's assume \( b = 1 \) (CH4). Then H: \( 2d = 4(1) \) → \( d = 2 \). O: \( d = 2a \) → \( 2 = 2a \) → \( a = 1 \). C: \( c = 1 + 1 = 2 \). So equation: \( \ce{2C + 2H2O -> CH4 + CO2} \). Wait, no, when \( a = 1 \) (CO2), \( b = 1 \) (CH4), \( c = 2 \) (C), \( d = 2 \) (H2O). Now check O: left \( 2\times1 = 2 \), right \( 2\times1 = 2 \). H: left \( 2\times2 = 4 \), right \( 4 \). C: left 2, right 1 + 1 = 2. So the coefficient of \( \ce{CO2} \) is 1? Wait, no, wait the original equation is \( \ce{C + H2O -> CH4 + CO2} \). Let's use variables. Let the coefficient of \( \ce{CO2} \) be \( x \), \( \ce{CH4} \) be \( y \), \( \ce{C} \) be \( z \), \( \ce{H2O} \) be \( w \).

C: \( z=y + x \)

H: \( 2w = 4y \) → \( w = 2y \)

O: \( w = 2x \)

Substitute \( w = 2y \) into \( w = 2x \): \( 2y=2x \) → \( y = x \)

From C: \( z=y + x = 2x \)

Let's take \( x = 1 \), then \( y = 1 \), \( z = 2 \), \( w = 2 \)

So the equation is \( \ce{2C + 2H2O -> CH4 + CO2} \). Wait, but the question is about the coefficient of \( \ce{CO2} \), which is 1. Wait, no, let's check with \( x = 1 \). If \( x = 1 \), \( y = 1 \), \( z = 2 \), \( w = 2 \). So the equation is \( 2\ce{C}+2\ce{H2O}=\ce{CH4}+\ce{CO2} \). Wait, but when we count atoms:

Left: C: 2, H: 4, O: 2

Right: C: 1 + 1 = 2, H: 4, O: 2. Yes, balanced. So the coefficient of \( \ce{CO2} \) is 1.

Wait, maybe a simpler way: Let's balance C first. There are 2 C on the right (1 in \( \ce{CH4} \), 1 in \( \ce{CO2} \)), so we need 2 C on the left: \( 2\ce{C}+\ce{H2O}\to\ce{CH4}+\ce{CO2} \). Now balance H: \( \ce{CH4} \) has 4 H, so \( \ce{H2O} \) needs 2 molecules (4 H): \( 2\ce{C}+2\ce{H2O}\to\ce{CH4}+\ce{CO2} \). Now O: 2 \( \ce{H2O} \) has 2 O, and \( \ce{CO2} \) has 2 O. Balanced. So coefficient of \( \ce{CO2} \) is 1.

Step1: Balance Nitrogen atoms

Left: \( \ce{NH3} \) has 1 N. Right: \( \ce{NO2} \) has 1 N. Let coefficient of \( \ce{NH3} \) be \( a \) and \( \ce{NO2} \) be \( a \): \( \ce{aNH3 + O2 -> aNO2 + H2O} \).

Step2: Balance Hydrogen atoms

Left: \( 3a \) H. Right: \( 2b \) H (from \( b \) \( \ce{H2O} \)). So \( 3a = 2b \). Let's take \( a = 4 \), then \( b = 6 \) (since \( 3\times4 = 12 \), \( 2\times6 = 12 \)).

Step3: Balance Oxygen atoms

Right: \( a\times2 + b\times1 = 4\times2+6\times1 = 8 + 6 = 14 \) O. Left: \( \ce{O2} \) has \( 2c \) O. So \( 2c = 14 \) → \( c = 7 \). Now the equation: \( \ce{4NH3 + 7O2 -> 4NO2 + 6H2O} \). The coefficient for \( \ce{H2O} \) is 6.

Answer:

2

Level 2: Balancing \( \boldsymbol{\ce{C + H2O -> CH4 + CO2}} \)