How Many Grams Of $H_3PO_4$ Are Produced In The Reaction Starting From 5.60 G Of $Ca_3(PO_4)_2$?

$$$$

Introduction

In this article, we will explore the production of $H_3PO_4$ (phosphoric acid) from $Ca_3(PO_4)_2$ (tricalcium phosphate) through a chemical reaction. The reaction involves the acidification of $Ca_3(PO_4)_2$ with $HCl$ (hydrochloric acid) to produce $H_3PO_4$ and $CaCl_2$ (calcium chloride). We will calculate the number of grams of $H_3PO_4$ produced in the reaction starting from 5.60 g of $Ca_3(PO_4)_2$.

Chemical Reaction

The chemical reaction between $Ca_3(PO_4)_2$ and $HCl$ can be represented by the following equation:

$$Ca_3(PO_4)_2 + 6HCl \rightarrow 2H_3PO_4 + 3CaCl_2$$

In this reaction, one mole of $Ca_3(PO_4)_2$ reacts with six moles of $HCl$ to produce two moles of $H_3PO_4$ and three moles of $CaCl_2$.

Molar Mass Calculations

To calculate the number of grams of $H_3PO_4$ produced, we need to calculate the molar mass of $Ca_3(PO_4)_2$ and $H_3PO_4$. The molar mass of a compound is the sum of the atomic masses of its constituent elements.

The atomic masses of the elements are:

• $Ca$: 40.08 g/mol
• $P$: 30.97 g/mol
• $O$: 16.00 g/mol
• $H$: 1.01 g/mol
• $Cl$: 35.45 g/mol

The molar mass of $Ca_3(PO_4)_2$ is:

$$3(40.08) + 2(30.97) + 8(16.00) = 310.18 g/mol$$

The molar mass of $H_3PO_4$ is:

$$3(1.01) + 30.97 + 4(16.00) = 98.00 g/mol$$

Calculating the Number of Moles of $Ca_3(PO_4)_2$

To calculate the number of moles of $Ca_3(PO_4)_2$, we can use the formula:

$$n = \frac{m}{M}$$

where $n$ is the number of moles, $m$ is the mass of the substance, and $M$ is the molar mass.

Given that the mass of $Ca_3(PO_4)_2$ is 5.60 g, we can calculate the number of moles as follows:

$$n = \frac{5.60 g}{310.18 g/mol} = 0.0181 mol$$

Calculating the Number of Moles of $H_3PO_4$ Produced

From the chemical reaction equation, we can see that two moles of $H_3PO_4$ are produced for every one mole of $Ca_3(PO_4)_2$ reacted. Therefore, the number of moles of $H_3PO_4$ produced is:

$$n = 2 \times 0.0181 mol = 0.0362 mol$$

Calculating the Mass of $H_3PO_4$ Produced

To calculate the mass of $H_3PO_4$ produced, we can use the formula:

$$m = n \times M$$

where $m$ is the mass, $n$ is the number of moles, and $M$ is the molar mass.

Given that the number of moles of $H_3PO_4$ produced is 0.0362 mol and the molar mass of $H_3PO_4$ is 98.00 g/mol, we can calculate the mass as follows:

$$m = 0.0362 mol \times 98.00 g/mol = 3.55 g$$

Conclusion

In this article, we calculated the number of grams of $H_3PO_4$ produced in the reaction starting from 5.60 g of $Ca_3(PO_4)_2$. We first calculated the molar mass of $Ca_3(PO_4)_2$ and $H_3PO_4$, then calculated the number of moles of $Ca_3(PO_4)_2$ and $H_3PO_4$ produced, and finally calculated the mass of $H_3PO_4$ produced. The result shows that 3.55 g of $H_3PO_4$ are produced in the reaction.

References

• CRC Handbook of Chemistry and Physics, 97th Edition
• NIST Chemistry WebBook
• Wikipedia: Phosphoric acid
• Wikipedia: Tricalcium phosphate
  

$$$$

Q: What is the chemical reaction between $Ca_3(PO_4)_2$ and $HCl$?

A: The chemical reaction between $Ca_3(PO_4)_2$ and $HCl$ can be represented by the following equation:

$$Ca_3(PO_4)_2 + 6HCl \rightarrow 2H_3PO_4 + 3CaCl_2$$

Q: What is the molar mass of $Ca_3(PO_4)_2$ and $H_3PO_4$?

A: The molar mass of $Ca_3(PO_4)_2$ is 310.18 g/mol, and the molar mass of $H_3PO_4$ is 98.00 g/mol.

Q: How many moles of $Ca_3(PO_4)_2$ are required to produce 1 mole of $H_3PO_4$?

A: From the chemical reaction equation, we can see that 1 mole of $Ca_3(PO_4)_2$ is required to produce 2 moles of $H_3PO_4$.

Q: What is the mass of $H_3PO_4$ produced when 5.60 g of $Ca_3(PO_4)_2$ is reacted with $HCl$?

A: The mass of $H_3PO_4$ produced is 3.55 g.

Q: What are the by-products of the reaction between $Ca_3(PO_4)_2$ and $HCl$?

A: The by-products of the reaction are $CaCl_2$ and $H_2O$.

Q: Can the reaction between $Ca_3(PO_4)_2$ and $HCl$ be used to produce other compounds?

A: Yes, the reaction between $Ca_3(PO_4)_2$ and $HCl$ can be used to produce other compounds, such as $Ca(OH)_2$ and $H_3PO_4$.

Q: What are the applications of $H_3PO_4$?

A: $H_3PO_4$ has many applications, including:

• Food industry: as a preservative and acidity regulator
• Pharmaceutical industry: as a solvent and intermediate
• Industrial cleaning: as a degreaser and cleaning agent
• Water treatment: as a coagulant and flocculant

Q: What are the safety precautions when handling $Ca_3(PO_4)_2$ and $HCl$?

A: When handling $Ca_3(PO_4)_2$ and $HCl$, it is essential to wear protective gear, including gloves, goggles, and a face mask. The reaction between $Ca_3(PO_4)_2$ and $HCl$ can produce heat and release toxic fumes, so it is crucial to handle the substances in a well-ventilated area and follow proper safety protocols.

Q: Can the reaction between $Ca_3(PO_4)_2$ and $HCl$ be scaled up for industrial production?

A: Yes, the reaction between $Ca_3(PO_4)_2$ and $HCl$ can be scaled up for industrial production. However, it is essential to consider the safety and environmental implications of large-scale production and to implement proper safety protocols and waste management systems.

Q: What are the environmental implications of the reaction between $Ca_3(PO_4)_2$ and $HCl$?

A: The reaction between $Ca_3(PO_4)_2$ and $HCl$ can produce waste products, including $CaCl_2$ and $H_3PO_4$, which can have environmental implications if not properly managed. It is essential to implement proper waste management systems and to consider the environmental implications of large-scale production.

Q: Can the reaction between $Ca_3(PO_4)_2$ and $HCl$ be used to produce $H_3PO_4$ in a more efficient and cost-effective manner?

A: Yes, the reaction between $Ca_3(PO_4)_2$ and $HCl$ can be optimized to produce $H_3PO_4$ in a more efficient and cost-effective manner. This can be achieved by optimizing the reaction conditions, such as temperature, pressure, and reactant ratios, and by implementing proper safety protocols and waste management systems.