Math Snap

STEP 1

1. The chemical equation given is:
$$ \mathrm{HF} + \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightarrow \mathrm{CaF}_{2} + \mathrm{H}_{3}\left(\mathrm{PO}_{4}\right)
\] 2. We need to balance the equation by ensuring the same number of each type of atom on both sides of the equation.

STEP 2

1. Identify the number of each type of atom in the reactants and products.
2. Balance the equation by adjusting coefficients.
3. Verify that the equation is balanced.

STEP 3

Identify the number of each type of atom in the reactants and products.
- Reactants:
- $\mathrm{HF}$: 1 H, 1 F
- $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$: 3 Ca, 2 P, 8 O
- Products:
- $\mathrm{CaF}_{2}$: 1 Ca, 2 F
- $\mathrm{H}_{3}\left(\mathrm{PO}_{4}\right)$: 3 H, 1 P, 4 O

STEP 4

Balance the equation by adjusting coefficients.
- Start by balancing the calcium ($\mathrm{Ca}$) atoms:
- Reactants have 3 Ca from $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$.
- Products have 1 Ca in $\mathrm{CaF}_{2}$.
- Place a coefficient of 3 in front of $\mathrm{CaF}_{2}$:
$$ \mathrm{HF} + \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightarrow 3\mathrm{CaF}_{2} + \mathrm{H}_{3}\left(\mathrm{PO}_{4}\right)
\] - Balance the fluorine ($\mathrm{F}$) atoms:
- Products now have 6 F (from 3 $\mathrm{CaF}_{2}$).
- Reactants have 1 F in $\mathrm{HF}$.
- Place a coefficient of 6 in front of $\mathrm{HF}$:
$$ 6\mathrm{HF} + \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightarrow 3\mathrm{CaF}_{2} + \mathrm{H}_{3}\left(\mathrm{PO}_{4}\right)
\] - Balance the hydrogen ($\mathrm{H}$) atoms:
- Reactants now have 6 H (from 6 $\mathrm{HF}$).
- Products have 3 H in $\mathrm{H}_{3}\left(\mathrm{PO}_{4}\right)$.
- Place a coefficient of 2 in front of $\mathrm{H}_{3}\left(\mathrm{PO}_{4}\right)$:
$$ 6\mathrm{HF} + \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightarrow 3\mathrm{CaF}_{2} + 2\mathrm{H}_{3}\left(\mathrm{PO}_{4}\right)
\] - Balance the phosphorus ($\mathrm{P}$) and oxygen ($\mathrm{O}$) atoms:
- Reactants have 2 P and 8 O from $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$.
- Products have 2 P and 8 O from 2 $\mathrm{H}_{3}\left(\mathrm{PO}_{4}\right)$.

Verify that the equation is balanced.
- Reactants: 6 H, 6 F, 3 Ca, 2 P, 8 O
- Products: 6 H, 6 F, 3 Ca, 2 P, 8 O
The equation is balanced.
The correct balanced coefficients are: $\boxed{6, 1, 3, 2}$