Math / Algebra

QuestionBalance this equation: $-\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})+\_\mathrm{HF}(\ell) \rightarrow-\mathrm{BF}_{3}(\mathrm{~s})+\ldots \mathrm{H}_{2} \mathrm{O}(\ell)$
Though you would not normally do so, enter the coefficient of "1" if needed. $\square$ $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})$ $\square$ HF( $\ell$ ) $\square$ $\mathrm{BF}_{3}(\mathrm{~s})$ $\square$ $\mathrm{H}_{2} \mathrm{O}(\ell)$

Studdy Solution

STEP 1

What is this asking? We need to find the smallest whole number coefficients that balance the given chemical equation, making sure the number of atoms of each element is the same on both sides. Watch out! Don't forget that the coefficient multiplies every atom in the molecule.
Also, a missing coefficient is the same as a coefficient of 1!

STEP 2

1. Balance the Boron (B) atoms.
2. Balance the Oxygen (O) atoms.
3. Balance the Hydrogen (H) atoms.
4. Balance the Fluorine (F) atoms.
5. Final Check!

STEP 3

We've got $\mathrm{B}_{2} \mathrm{O}_{3}$ on the left with two Boron atoms, and $\mathrm{BF}_{3}$ on the right with only one Boron atom.
To balance the Boron atoms, we'll put a coefficient of 2 in front of $\mathrm{BF}_{3}$ .

STEP 4

Now our equation looks like this: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{HF}(\ell) \rightarrow \mathbf{2} \mathrm{BF}_{3}(\mathrm{~s}) + \mathrm{H}_{2} \mathrm{O}(\ell)$ Woohoo! Boron is balanced!

STEP 5

We see three Oxygen atoms on the left side with $\mathrm{B}_{2} \mathrm{O}_{3}$ and one Oxygen atom on the right side with $\mathrm{H}_{2} \mathrm{O}$ .
Let's add a coefficient of 3 in front of $\mathrm{H}_{2} \mathrm{O}$ to balance the Oxygen atoms.

STEP 6

Our updated equation is: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{HF}(\ell) \rightarrow 2 \mathrm{BF}_{3}(\mathrm{~s}) + \mathbf{3} \mathrm{H}_{2} \mathrm{O}(\ell)$ Oxygen is now balanced!

STEP 7

We have one Hydrogen atom on the left side with $\mathrm{HF}$ and six Hydrogen atoms on the right side with $3\mathrm{H}_{2}\mathrm{O}$ ( $3 \cdot 2 = 6$ ).
To balance the Hydrogen atoms, we'll put a coefficient of 6 in front of $\mathrm{HF}$ .

STEP 8

The equation now looks like this: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathbf{6} \mathrm{HF}(\ell) \rightarrow 2 \mathrm{BF}_{3}(\mathrm{~s}) + 3 \mathrm{H}_{2} \mathrm{O}(\ell)$ Hydrogen is balanced!

STEP 9

We have six Fluorine atoms on the left side with $6\mathrm{HF}$ ( $6 \cdot 1 = 6$ ) and six Fluorine atoms on the right side with $2\mathrm{BF}_{3}$ ( $2 \cdot 3 = 6$ ).
The Fluorine atoms are already balanced!

STEP 10

Let's double-check: * Boron (B): 2 on both sides. * Oxygen (O): 3 on both sides. * Hydrogen (H): 6 on both sides. * Fluorine (F): 6 on both sides.
Everything is perfectly balanced!

STEP 11

$1$ $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})$ $6$ $\mathrm{HF}(\ell)$ $2$ $\mathrm{BF}_{3}(\mathrm{~s})$ $3$ $\mathrm{H}_{2} \mathrm{O}(\ell)$

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Studdy Solution

STEP 1

STEP 2

1. Balance the Boron (B) atoms.
2. Balance the Oxygen (O) atoms.
3. Balance the Hydrogen (H) atoms.
4. Balance the Fluorine (F) atoms.
5. Final Check!

STEP 3

We've got $\mathrm{B}_{2} \mathrm{O}_{3}$ on the left with two Boron atoms, and $\mathrm{BF}_{3}$ on the right with only one Boron atom.
To balance the Boron atoms, we'll put a coefficient of 2 in front of $\mathrm{BF}_{3}$ .

STEP 4

Now our equation looks like this: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{HF}(\ell) \rightarrow \mathbf{2} \mathrm{BF}_{3}(\mathrm{~s}) + \mathrm{H}_{2} \mathrm{O}(\ell)$ Woohoo! Boron is balanced!

STEP 5

We see three Oxygen atoms on the left side with $\mathrm{B}_{2} \mathrm{O}_{3}$ and one Oxygen atom on the right side with $\mathrm{H}_{2} \mathrm{O}$ .
Let's add a coefficient of 3 in front of $\mathrm{H}_{2} \mathrm{O}$ to balance the Oxygen atoms.

STEP 6

Our updated equation is: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{HF}(\ell) \rightarrow 2 \mathrm{BF}_{3}(\mathrm{~s}) + \mathbf{3} \mathrm{H}_{2} \mathrm{O}(\ell)$ Oxygen is now balanced!

STEP 7

We have one Hydrogen atom on the left side with $\mathrm{HF}$ and six Hydrogen atoms on the right side with $3\mathrm{H}_{2}\mathrm{O}$ ( $3 \cdot 2 = 6$ ).
To balance the Hydrogen atoms, we'll put a coefficient of 6 in front of $\mathrm{HF}$ .

STEP 8

The equation now looks like this: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathbf{6} \mathrm{HF}(\ell) \rightarrow 2 \mathrm{BF}_{3}(\mathrm{~s}) + 3 \mathrm{H}_{2} \mathrm{O}(\ell)$ Hydrogen is balanced!

STEP 9

We have six Fluorine atoms on the left side with $6\mathrm{HF}$ ( $6 \cdot 1 = 6$ ) and six Fluorine atoms on the right side with $2\mathrm{BF}_{3}$ ( $2 \cdot 3 = 6$ ).
The Fluorine atoms are already balanced!

STEP 10

Let's double-check: * Boron (B): 2 on both sides. * Oxygen (O): 3 on both sides. * Hydrogen (H): 6 on both sides. * Fluorine (F): 6 on both sides.
Everything is perfectly balanced!

STEP 11

$1$ $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})$ $6$ $\mathrm{HF}(\ell)$ $2$ $\mathrm{BF}_{3}(\mathrm{~s})$ $3$ $\mathrm{H}_{2} \mathrm{O}(\ell)$

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Studdy Solution

STEP 1

STEP 2

1. Balance the Boron (B) atoms.2. Balance the Oxygen (O) atoms.3. Balance the Hydrogen (H) atoms.4. Balance the Fluorine (F) atoms.5. Final Check!

STEP 3

We've got B2O3\mathrm{B}_{2} \mathrm{O}_{3}B2​O3​ on the left with **two** Boron atoms, and BF3\mathrm{BF}_{3}BF3​ on the right with only **one** Boron atom.To balance the Boron atoms, we'll put a **coefficient of 2** in front of BF3\mathrm{BF}_{3}BF3​.

STEP 4

Now our equation looks like this: B2O3( s)+HF(ℓ)→2BF3( s)+H2O(ℓ) \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{HF}(\ell) \rightarrow \mathbf{2} \mathrm{BF}_{3}(\mathrm{~s}) + \mathrm{H}_{2} \mathrm{O}(\ell) B2​O3​( s)+HF(ℓ)→2BF3​( s)+H2​O(ℓ) Woohoo! Boron is balanced!

STEP 5

We see **three** Oxygen atoms on the left side with B2O3\mathrm{B}_{2} \mathrm{O}_{3}B2​O3​ and **one** Oxygen atom on the right side with H2O\mathrm{H}_{2} \mathrm{O}H2​O.Let's add a **coefficient of 3** in front of H2O\mathrm{H}_{2} \mathrm{O}H2​O to balance the Oxygen atoms.

STEP 6

Our updated equation is: B2O3( s)+HF(ℓ)→2BF3( s)+3H2O(ℓ) \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{HF}(\ell) \rightarrow 2 \mathrm{BF}_{3}(\mathrm{~s}) + \mathbf{3} \mathrm{H}_{2} \mathrm{O}(\ell) B2​O3​( s)+HF(ℓ)→2BF3​( s)+3H2​O(ℓ) Oxygen is now balanced!

STEP 7

We have **one** Hydrogen atom on the left side with HF\mathrm{HF}HF and **six** Hydrogen atoms on the right side with 3H2O3\mathrm{H}_{2}\mathrm{O}3H2​O (3⋅2=63 \cdot 2 = 63⋅2=6).To balance the Hydrogen atoms, we'll put a **coefficient of 6** in front of HF\mathrm{HF}HF.

STEP 8

The equation now looks like this: B2O3( s)+6HF(ℓ)→2BF3( s)+3H2O(ℓ) \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathbf{6} \mathrm{HF}(\ell) \rightarrow 2 \mathrm{BF}_{3}(\mathrm{~s}) + 3 \mathrm{H}_{2} \mathrm{O}(\ell) B2​O3​( s)+6HF(ℓ)→2BF3​( s)+3H2​O(ℓ) Hydrogen is balanced!

STEP 9

We have **six** Fluorine atoms on the left side with 6HF6\mathrm{HF}6HF (6⋅1=66 \cdot 1 = 66⋅1=6) and **six** Fluorine atoms on the right side with 2BF32\mathrm{BF}_{3}2BF3​ (2⋅3=62 \cdot 3 = 62⋅3=6).The Fluorine atoms are already balanced!

STEP 10

Let's double-check: * **Boron (B):** 2 on both sides. * **Oxygen (O):** 3 on both sides. * **Hydrogen (H):** 6 on both sides. * **Fluorine (F):** 6 on both sides.Everything is perfectly balanced!

STEP 11

111 B2O3( s)\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})B2​O3​( s) 666 HF(ℓ)\mathrm{HF}(\ell)HF(ℓ) 222 BF3( s)\mathrm{BF}_{3}(\mathrm{~s})BF3​( s) 333 H2O(ℓ)\mathrm{H}_{2} \mathrm{O}(\ell)H2​O(ℓ)

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1. Balance the Boron (B) atoms.
2. Balance the Oxygen (O) atoms.
3. Balance the Hydrogen (H) atoms.
4. Balance the Fluorine (F) atoms.
5. Final Check!

We've got $\mathrm{B}_{2} \mathrm{O}_{3}$ on the left with two Boron atoms, and $\mathrm{BF}_{3}$ on the right with only one Boron atom.
To balance the Boron atoms, we'll put a coefficient of 2 in front of $\mathrm{BF}_{3}$ .

Now our equation looks like this: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{HF}(\ell) \rightarrow \mathbf{2} \mathrm{BF}_{3}(\mathrm{~s}) + \mathrm{H}_{2} \mathrm{O}(\ell)$ Woohoo! Boron is balanced!

We see three Oxygen atoms on the left side with $\mathrm{B}_{2} \mathrm{O}_{3}$ and one Oxygen atom on the right side with $\mathrm{H}_{2} \mathrm{O}$ .
Let's add a coefficient of 3 in front of $\mathrm{H}_{2} \mathrm{O}$ to balance the Oxygen atoms.

Our updated equation is: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{HF}(\ell) \rightarrow 2 \mathrm{BF}_{3}(\mathrm{~s}) + \mathbf{3} \mathrm{H}_{2} \mathrm{O}(\ell)$ Oxygen is now balanced!

We have one Hydrogen atom on the left side with $\mathrm{HF}$ and six Hydrogen atoms on the right side with $3\mathrm{H}_{2}\mathrm{O}$ ( $3 \cdot 2 = 6$ ).
To balance the Hydrogen atoms, we'll put a coefficient of 6 in front of $\mathrm{HF}$ .

The equation now looks like this: $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathbf{6} \mathrm{HF}(\ell) \rightarrow 2 \mathrm{BF}_{3}(\mathrm{~s}) + 3 \mathrm{H}_{2} \mathrm{O}(\ell)$ Hydrogen is balanced!

We have six Fluorine atoms on the left side with $6\mathrm{HF}$ ( $6 \cdot 1 = 6$ ) and six Fluorine atoms on the right side with $2\mathrm{BF}_{3}$ ( $2 \cdot 3 = 6$ ).
The Fluorine atoms are already balanced!

Let's double-check: * Boron (B): 2 on both sides. * Oxygen (O): 3 on both sides. * Hydrogen (H): 6 on both sides. * Fluorine (F): 6 on both sides.
Everything is perfectly balanced!

$1$ $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})$ $6$ $\mathrm{HF}(\ell)$ $2$ $\mathrm{BF}_{3}(\mathrm{~s})$ $3$ $\mathrm{H}_{2} \mathrm{O}(\ell)$