Question

fill in the blank 2 points
balance the following equation:
3
sio₂ + 2
hf --> 4
sif₄ + 2
h₂o

Explanation:

Step1: Identify atoms to balance

We need to balance the chemical equation \( \text{SiO}_2 + \text{HF} ightarrow \text{SiF}_4 + \text{H}_2\text{O} \). Let's list the number of each atom on both sides:
- Si: 1 on left, 1 on right (balanced for now)
- O: 2 on left, 1 on right (in \( \text{H}_2\text{O} \))
- F: 1 on left (in HF), 4 on right (in \( \text{SiF}_4 \))
- H: 1 on left (in HF), 2 on right (in \( \text{H}_2\text{O} \))

Step2: Balance F atoms

To balance F, since there are 4 F on the right (in \( \text{SiF}_4 \)), we need 4 HF on the left. So put a coefficient of 4 in front of HF: \( \text{SiO}_2 + 4\text{HF} ightarrow \text{SiF}_4 + \text{H}_2\text{O} \)

Step3: Balance H and O atoms

Now, with 4 HF, we have 4 H on the left. On the right, \( \text{H}_2\text{O} \) has 2 H per molecule, so we need 2 \( \text{H}_2\text{O} \) to balance H (4 H total). Let's check O: left has 2 (in \( \text{SiO}_2 \)), right has 2 (in 2 \( \text{H}_2\text{O} \)). Now the equation is balanced: \( \text{SiO}_2 + 4\text{HF} ightarrow \text{SiF}_4 + 2\text{H}_2\text{O} \)

Wait, but the given partial equation has some coefficients. Wait, the original problem's partial equation: Let's re-express. The equation is \( 3\text{SiO}_2 + 2\text{HF} ightarrow 2\text{SiF}_4 + 2\text{H}_2\text{O} \)? No, wait, no—wait the user's image: Wait, the correct balancing: Wait, no, let's do it properly.

Wait, the correct balanced equation for \( \text{SiO}_2 + \text{HF} ightarrow \text{SiF}_4 + \text{H}_2\text{O} \) is:

1 \( \text{SiO}_2 \) + 4 \( \text{HF} \) → 1 \( \text{SiF}_4 \) + 2 \( \text{H}_2\text{O} \)

But the given problem has some pre-filled coefficients: "3 \( \text{SiO}_2 \) + 2 \( \text{HF} \) → 2 \( \text{SiF}_4 \) + 2 \( \text{H}_2\text{O} \)"? No, that can't be. Wait, maybe the problem is a different approach. Wait, let's check the atoms again.

Wait, maybe the equation is \( \text{SiO}_2 + \text{HF} ightarrow \text{SiF}_4 + \text{H}_2\text{O} \). Let's balance step by step:

- Si: 1 on each side.
- F: 4 on right (in \( \text{SiF}_4 \)), so need 4 HF: \( \text{SiO}_2 + 4\text{HF} ightarrow \text{SiF}_4 + \text{H}_2\text{O} \)
- H: 4 on left (4 HF), so need 2 \( \text{H}_2\text{O} \) (2×2=4 H): \( \text{SiO}_2 + 4\text{HF} ightarrow \text{SiF}_4 + 2\text{H}_2\text{O} \)
- O: 2 on left ( \( \text{SiO}_2 \) ), 2 on right (2 \( \text{H}_2\text{O} \) ), balanced.

But the given problem has "3 \( \text{SiO}_2 \)"? Wait, maybe the user's problem has a typo, but assuming the standard balancing, the coefficient for HF should be 4, for \( \text{H}_2\text{O} \) should be 2, and for \( \text{SiF}_4 \) should be 1, \( \text{SiO}_2 \) 1. But the given partial coefficients: Let's see the user's image:

The blanks: 3 (for \( \text{SiO}_2 \)), 2 (for \( \text{H}_2\text{O} \)), 2 (for HF), 4 (for \( \text{SiF}_4 \))? Wait, no, let's re-express the equation as per the blanks. Wait, the equation is:

\( 3\text{SiO}_2 + \boldsymbol{4}\text{HF} ightarrow \boldsymbol{3}\text{SiF}_4 + 2\text{H}_2\text{O} \)? No, that's not balanced. Wait, no—wait the correct balancing is \( \text{SiO}_2 + 4\text{HF} = \text{SiF}_4 + 2\text{H}_2\text{O} \). So if the given equation has 3 \( \text{SiO}_2 \), then we need to scale. Wait, 3 \( \text{SiO}_2 \) would need 12 HF, 3 \( \text{SiF}_4 \), and 6 \( \text{H}_2\text{O} \). But the user's blanks: 3 ( \( \text{SiO}_2 \) ), 2 ( \( \text{H}_2\text{O} \) ), 2 (HF), 4 ( \( \text{SiF}_4 \) ). Wait, maybe the problem has a different setup. Wait, perhaps the intended equation is \( 3\text{SiO}_2 + 12\text{HF} = 3\text{SiF}_4 + 6\text{H}_2\text{O} \), but simplified by dividing by 3: \( \text{SiO}_2 + 4\text{HF} = \text{SiF}_4 + 2\text{H}_2\text{O} \). But the user's blanks: the HF coefficient is 2? No, that's wrong. Wait, maybe the user made a typo, but according to the standard balancing, the coefficient for HF is 4, for \( \text{H}_2\text{O} \) is 2, for \( \text{SiF}_4 \) is 1, and \( \text{SiO}_2 \) is 1. But the given partial equation has 3 \( \text{SiO}_2 \), 2 \( \text{H}_2\text{O} \), 2 HF, 4 \( \text{SiF}_4 \). Wait, no—let's check the atoms:

Left: 3 Si (3 \( \text{SiO}_2 \)), 6 O (3×2), 2 H (2 HF), 2 F (2 HF)

Right: 4 Si (4 \( \text{SiF}_4 \))—no, that's not balanced. So there's a mistake. Wait, the correct balancing is \( \text{SiO}_2 + 4\text{HF} = \text{SiF}_4 + 2\text{H}_2\text{O} \). So the missing coefficient for HF is 4, for \( \text{SiF}_4 \) is 1, but the user's blanks have 2 for HF and 4 for \( \text{SiF}_4 \), 3 for \( \text{SiO}_2 \), 2 for \( \text{H}_2\text{O} \). Wait, maybe the equation is \( 3\text{SiO}_2 + 12\text{HF} = 3\text{SiF}_4 + 6\text{H}_2\text{O} \), and then dividing by 3: \( \text{SiO}_2 + 4\text{HF} = \text{SiF}_4 + 2\text{H}_2\text{O} \). So if the blanks are for HF: 4, \( \text{SiF}_4 \): 1, but the user's blanks have 2 for HF and 4 for \( \text{SiF}_4 \). I think there's a mistake in the problem's partial coefficients, but the correct coefficient for HF in the balanced equation (when \( \text{SiO}_2 \) is 1) is 4, for \( \text{H}_2\text{O} \) is 2, for \( \text{SiF}_4 \) is 1. So if the given equation has 3 \( \text{SiO}_2 \), then HF is 12, \( \text{SiF}_4 \) is 3, \( \text{H}_2\text{O} \) is 6. But the user's blanks: 3 ( \( \text{SiO}_2 \) ), 2 ( \( \text{H}_2\text{O} \) ), 2 (HF), 4 ( \( \text{SiF}_4 \) ). Wait, maybe the intended answer is 4 for HF (the blank for HF), 3 for \( \text{SiF}_4 \) (if \( \text{SiO}_2 \) is 3), and 6 for \( \text{H}_2\text{O} \), but the user's blanks have 2 for \( \text{H}_2\text{O} \). I think there's a confusion, but the correct coefficient for HF in the standard balanced equation (1 \( \text{SiO}_2 \)) is 4. So the blank for HF should be 4.

Answer:

The coefficient for HF should be 4, so the filled blank for HF is 4. The balanced equation (with the given partial coefficients adjusted) would be \( 3\text{SiO}_2 + 12\text{HF}
ightarrow 3\text{SiF}_4 + 6\text{H}_2\text{O} \), but if we take the standard 1:4:1:2, then HF is 4. So the answer for the HF blank is 4.