3. phosphoric acid can be prepared from elemental phosphorus, oxygen, and water by the following series of reactions. be sure to balance the following reactions.
$mathrm{p}_{4}+mathrm{o}_{2} o mathrm{p}_{2} mathrm{o}_{5}$
$mathrm{p}_{2} mathrm{o}_{5}+mathrm{h}_{2} mathrm{o} o mathrm{h}_{3} mathrm{po}_{4}$
what volume of acid is produced from reacting 1.250 grams of phosphorus with sufficient oxygen and water? the specific gravity of phosphoric acid is 1.18.
4. potassium chlorate decomposes on heating to form potassium chloride and oxygen.
a. write the balanced equation for the reaction.
b. if 4.904 grams of potassium chlorate, with a 75.00% purity, is heated, what mass of oxygen may be produced?
c. 1.208 grams of oxygen are experimentally obtained. what is the percent yield?

Question

3. phosphoric acid can be prepared from elemental phosphorus, oxygen, and water by the following series of reactions. be sure to balance the following reactions.
$mathrm{p}_{4}+mathrm{o}_{2} 	o mathrm{p}_{2} mathrm{o}_{5}$
$mathrm{p}_{2} mathrm{o}_{5}+mathrm{h}_{2} mathrm{o} 	o mathrm{h}_{3} mathrm{po}_{4}$
what volume of acid is produced from reacting 1.250 grams of phosphorus with sufficient oxygen and water? the specific gravity of phosphoric acid is 1.18.
4. potassium chlorate decomposes on heating to form potassium chloride and oxygen.
a. write the balanced equation for the reaction.
b. if 4.904 grams of potassium chlorate, with a 75.00% purity, is heated, what mass of oxygen may be produced?
c. 1.208 grams of oxygen are experimentally obtained. what is the percent yield?

Accepted Answer

### 3. Balancing the reactions and calculating the volume of phosphoric acid
#### Balancing the first reaction: $P_4 + O_2
ightarrow P_2O_5$
- **Step 1: Balance the phosphorus atoms**
  - On the left - hand side, we have $P_4$ (4 phosphorus atoms), and on the right - hand side, we have $P_2O_5$. To balance the phosphorus atoms, we put a coefficient of $2$ in front of $P_2O_5$. So the equation becomes $P_4+O_2
ightarrow 2P_2O_5$.
- **Step 2: Balance the oxygen atoms**
  - Now, on the right - hand side, we have $2\times5 = 10$ oxygen atoms. So we put a coefficient of $5$ in front of $O_2$. The balanced equation is $P_4 + 5O_2=2P_2O_5$.

#### Balancing the second reaction: $P_2O_5+H_2O
ightarrow H_3PO_4$
- **Step 1: Balance the phosphorus atoms**
  - We have $2$ phosphorus atoms on the left - hand side ($P_2O_5$) and $1$ on the right - hand side ($H_3PO_4$). So we put a coefficient of $2$ in front of $H_3PO_4$. The equation becomes $P_2O_5 + H_2O
ightarrow 2H_3PO_4$.
- **Step 2: Balance the oxygen and hydrogen atoms**
  - On the right - hand side, we have $6$ hydrogen atoms ($2\times3$ in $2H_3PO_4$) and $8$ oxygen atoms ($2\times4$ in $2H_3PO_4$). On the left - hand side, we have $5$ oxygen atoms in $P_2O_5$. So we put a coefficient of $3$ in front of $H_2O$. The balanced equation is $P_2O_5+3H_2O = 2H_3PO_4$.

#### Calculating the volume of phosphoric acid
- **Step 1: Calculate the moles of $P_4$**
  - The molar mass of $P_4$ is $M = 4\times30.97\space g/mol=123.88\space g/mol$. The moles of $P_4$, $n(P_4)=\frac{m(P_4)}{M(P_4)}=\frac{1.250\space g}{123.88\space g/mol}\approx0.0101\space mol$.
- **Step 2: Use stoichiometry to find moles of $H_3PO_4$**
  - From the first reaction $P_4 + 5O_2=2P_2O_5$ and the second reaction $P_2O_5+3H_2O = 2H_3PO_4$, the mole ratio of $P_4$ to $H_3PO_4$ is $1:4$. So $n(H_3PO_4)=4n(P_4)=4\times0.0101\space mol = 0.0404\space mol$.
- **Step 3: Calculate the mass of $H_3PO_4$**
  - The molar mass of $H_3PO_4$ is $M = 3\times1 + 30.97+4\times16=97.97\space g/mol$. The mass of $H_3PO_4$, $m(H_3PO_4)=n(H_3PO_4)\times M(H_3PO_4)=0.0404\space mol\times97.97\space g/mol\approx3.96\space g$.
- **Step 4: Calculate the volume of $H_3PO_4$**
  - Specific gravity $=\frac{
ho_{H_3PO_4}}{
ho_{water}}$, and $
ho_{water} = 1\space g/mL$, so $
ho_{H_3PO_4}=1.18\space g/mL$. Using the formula $V=\frac{m}{
ho}$, $V=\frac{3.96\space g}{1.18\space g/mL}\approx3.36\space mL$.

### 4. Potassium chlorate decomposition
#### a. Balancing the equation
- **Step 1: Write the un - balanced equation**
  - The un - balanced equation is $KClO_3
ightarrow KCl + O_2$.
- **Step 2: Balance the oxygen atoms**
  - We have $3$ oxygen atoms on the left ($KClO_3$) and $2$ on the right ($O_2$). The least common multiple of $3$ and $2$ is $6$. So we put a coefficient of $2$ in front of $KClO_3$ and $3$ in front of $O_2$. The equation becomes $2KClO_3
ightarrow KCl+3O_2$.
- **Step 3: Balance the potassium and chlorine atoms**
  - We put a coefficient of $2$ in front of $KCl$. The balanced equation is $2KClO_3\xrightarrow{\Delta}2KCl + 3O_2$.

#### b. Calculating the mass of oxygen
- **Step 1: Calculate the mass of pure $KClO_3$**
  - The mass of pure $KClO_3$, $m_{pure}=4.904\space g\times0.7500 = 3.678\space g$.
- **Step 2: Calculate the moles of $KClO_3$**
  - The molar mass of $KClO_3$ is $M = 39.10+35.45 + 3\times16=122.55\space g/mol$. The moles of $KClO_3$, $n(KClO_3)=\frac{m_{pure}}{M}=\frac{3.678\space g}{122.55\space g/mol}\approx0.0300\space mol$.
- **Step 3: Use stoichiometry to find moles of $O_2$**
  - From the balanced equation $2KClO_3\xrightarrow{\Delta}2KCl + 3O_2$, the mole ratio of $KClO_3$ to $O_2$ is $2:3$. So $n(O_2)=\frac{3}{2}n(KClO_3)=\frac{3}{2}\times0.0300\space mol = 0.0450\space mol$.
- **Step 4: Calculate the mass of $O_2$**
  - The molar mass of $O_2$ is $M = 32\space g/mol$. The mass of $O_2$, $m(O_2)=n(O_2)\times M = 0.0450\space mol\times32\space g/mol = 1.44\space g$.

#### c. Calculating the percent yield
- **Step 1: Use the percent - yield formula**
  - The percent - yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$.
  - We know the actual yield $m_{actual}=1.208\space g$ and the theoretical yield $m_{theoretical}=1.44\space g$.
  - $\text{Percent Yield}=\frac{1.208\space g}{1.44\space g}\times 100\%\approx83.9\%$.

# Answer:
- **3. Balanced equations**:
  - $P_4 + 5O_2=2P_2O_5$
  - $P_2O_5+3H_2O = 2H_3PO_4$
  - Volume of $H_3PO_4\approx3.36\space mL$
- **4. a. Balanced equation**: $2KClO_3\xrightarrow{\Delta}2KCl + 3O_2$
- **4. b. Mass of $O_2$**: $1.44\space g$
- **4. c. Percent yield**: $\approx83.9\%$

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QUESTION IMAGE

Question

Explanation:

3. Balancing the reactions and calculating the volume of phosphoric acid

Balancing the first reaction: \(P_4 + O_2

Balancing the second reaction: \(P_2O_5+H_2O

Calculating the volume of phosphoric acid

4. Potassium chlorate decomposition

a. Balancing the equation

b. Calculating the mass of oxygen

Answer: