Calculating the Number of Moles of Sn

Oct 13, 2025 by leolavanderia.com.br 24 views

Understanding the Chemical Equation

The chemical equation provided shows the reaction between tin (Sn) and hydrogen fluoride (HF), resulting in the formation of tin fluoride (SnF2) and hydrogen gas (H2). The equation is as follows:

${ \text{Sn} + 2 \text{HF} \rightarrow \text{SnF}_2 + \text{H}_2 }$

This equation indicates that one mole of tin reacts with two moles of hydrogen fluoride to produce one mole of tin fluoride and one mole of hydrogen gas.

Calculating the Number of Moles of Sn

To calculate the number of moles of Sn, we need to know the molar mass of HF, which is given as $20.01 \, \text{g/mol}$ . We also need to know the molar mass of Sn, which is $118.71 \, \text{g/mol}$ .

Let's assume we have a certain amount of HF, and we want to calculate the number of moles of Sn that will react with it. We can start by writing the balanced chemical equation:

${ \text{Sn} + 2 \text{HF} \rightarrow \text{SnF}_2 + \text{H}_2 }$

From the equation, we can see that 1 mole of Sn reacts with 2 moles of HF. Therefore, if we have a certain amount of HF, we can calculate the number of moles of Sn that will react with it using the following equation:

${ \text{moles of Sn} = \frac{\text{moles of HF}}{2} }$

However, we are not given the number of moles of HF. Instead, we are given the molar mass of HF and the mass of HF that will react with Sn. Let's assume the mass of HF is $x \, \text{g}$ .

We can calculate the number of moles of HF using the following equation:

${ \text{moles of HF} = \frac{\text{mass of HF}}{\text{molar mass of HF}} }$

Substituting the given values, we get:

${ \text{moles of HF} = \frac{x \, \text{g}}{20.01 \, \text{g/mol}} }$

Now, we can substitute this expression into the equation for the number of moles of Sn:

${ \text{moles of Sn} = \frac{\frac{x \, \text{g}}{20.01 \, \text{g/mol}}}{2} }$

Simplifying the equation, we get:

${ \text{moles of Sn} = \frac{x \, \text{g}}{40.02 \, \text{g/mol}} }$

Calculating the Number of Moles of Sn Using the Given Mass of HF

To calculate the number of moles of Sn, we need to know the mass of HF that will react with Sn. Let's assume the mass of HF is $x \, \text{g}$ .

We can calculate the number of moles of HF using the following equation:

${ \text{moles of HF} = \frac{\text{mass of HF}}{\text{molar mass of HF}} }$

Substituting the given values, we get:

${ \text{moles of HF} = \frac{x \, \text{g}}{20.01 \, \text{g/mol}} }$

Now, we can substitute this expression into the equation for the number of moles of Sn:

${ \text{moles of Sn} = \frac{\frac{x \, \text{g}}{20.01 \, \text{g/mol}}}{2} }$

Simplifying the equation, we get:

${ \text{moles of Sn} = \frac{x \, \text{g}}{40.02 \, \text{g/mol}} }$

Example Calculation

Let's assume the mass of HF is $40.02 \, \text{g}$ . We can calculate the number of moles of HF using the following equation:

${ \text{moles of HF} = \frac{\text{mass of HF}}{\text{molar mass of HF}} }$

Substituting the given values, we get:

${ \text{moles of HF} = \frac{40.02 \, \text{g}}{20.01 \, \text{g/mol}} }$

Simplifying the equation, we get:

${ \text{moles of HF} = 2 \, \text{mol} }$

Now, we can substitute this expression into the equation for the number of moles of Sn:

${ \text{moles of Sn} = \frac{\frac{40.02 \, \text{g}}{20.01 \, \text{g/mol}}}{2} }$

Simplifying the equation, we get:

${ \text{moles of Sn} = 1 \, \text{mol} }$

Therefore, if we have $40.02 \, \text{g}$ of HF, we will need $1 \, \text{mol}$ of Sn to react with it.

Conclusion

In conclusion, we have calculated the number of moles of Sn that will react with a certain amount of HF. We have used the balanced chemical equation and the molar mass of HF to derive the equation for the number of moles of Sn. We have also provided an example calculation to illustrate the process.

The final answer is $\boxed{1}$

Q: What is the chemical equation between tin (Sn) and hydrogen fluoride (HF)?

A: The chemical equation between tin (Sn) and hydrogen fluoride (HF) is:

${ \text{Sn} + 2 \text{HF} \rightarrow \text{SnF}_2 + \text{H}_2 }$

This equation indicates that one mole of tin reacts with two moles of hydrogen fluoride to produce one mole of tin fluoride and one mole of hydrogen gas.

Q: What is the molar mass of hydrogen fluoride (HF)?

A: The molar mass of hydrogen fluoride (HF) is $20.01 \, \text{g/mol}$ .

Q: How many moles of Sn are required to react with a certain amount of HF?

A: To calculate the number of moles of Sn required to react with a certain amount of HF, we can use the following equation:

${ \text{moles of Sn} = \frac{\text{moles of HF}}{2} }$

However, we are not given the number of moles of HF. Instead, we are given the molar mass of HF and the mass of HF that will react with Sn. Let's assume the mass of HF is $x \, \text{g}$ .

We can calculate the number of moles of HF using the following equation:

${ \text{moles of HF} = \frac{\text{mass of HF}}{\text{molar mass of HF}} }$

Substituting the given values, we get:

${ \text{moles of HF} = \frac{x \, \text{g}}{20.01 \, \text{g/mol}} }$

Now, we can substitute this expression into the equation for the number of moles of Sn:

${ \text{moles of Sn} = \frac{\frac{x \, \text{g}}{20.01 \, \text{g/mol}}}{2} }$

Simplifying the equation, we get:

${ \text{moles of Sn} = \frac{x \, \text{g}}{40.02 \, \text{g/mol}} }$

Q: How many moles of Sn are required to react with 40.02 g of HF?

A: To calculate the number of moles of Sn required to react with 40.02 g of HF, we can use the equation:

${ \text{moles of Sn} = \frac{x \, \text{g}}{40.02 \, \text{g/mol}} }$

Substituting the given values, we get:

${ \text{moles of Sn} = \frac{40.02 \, \text{g}}{40.02 \, \text{g/mol}} }$

Simplifying the equation, we get:

${ \text{moles of Sn} = 1 \, \text{mol} }$

Therefore, if we have 40.02 g of HF, we will need 1 mol of Sn to react with it.

Q: What is the balanced chemical equation for the reaction between tin (Sn) and hydrogen fluoride (HF)?

A: The balanced chemical equation for the reaction between tin (Sn) and hydrogen fluoride (HF) is:

${ \text{Sn} + 2 \text{HF} \rightarrow \text{SnF}_2 + \text{H}_2 }$

This equation indicates that one mole of tin reacts with two moles of hydrogen fluoride to produce one mole of tin fluoride and one mole of hydrogen gas.

Q: What is the molar mass of tin (Sn)?

A: The molar mass of tin (Sn) is $118.71 \, \text{g/mol}$ .

Q: How many moles of HF are required to react with a certain amount of Sn?

A: To calculate the number of moles of HF required to react with a certain amount of Sn, we can use the following equation:

${ \text{moles of HF} = 2 \times \text{moles of Sn} }$

However, we are not given the number of moles of Sn. Instead, we are given the molar mass of Sn and the mass of Sn that will react with HF. Let's assume the mass of Sn is $x \, \text{g}$ .

We can calculate the number of moles of Sn using the following equation:

${ \text{moles of Sn} = \frac{\text{mass of Sn}}{\text{molar mass of Sn}} }$

Substituting the given values, we get:

${ \text{moles of Sn} = \frac{x \, \text{g}}{118.71 \, \text{g/mol}} }$

Now, we can substitute this expression into the equation for the number of moles of HF:

${ \text{moles of HF} = 2 \times \frac{x \, \text{g}}{118.71 \, \text{g/mol}} }$

Simplifying the equation, we get:

${ \text{moles of HF} = \frac{2x \, \text{g}}{118.71 \, \text{g/mol}} }$

Q: How many moles of HF are required to react with 118.71 g of Sn?

A: To calculate the number of moles of HF required to react with 118.71 g of Sn, we can use the equation:

${ \text{moles of HF} = \frac{2x \, \text{g}}{118.71 \, \text{g/mol}} }$

Substituting the given values, we get:

${ \text{moles of HF} = \frac{2 \times 118.71 \, \text{g}}{118.71 \, \text{g/mol}} }$

Simplifying the equation, we get:

${ \text{moles of HF} = 2 \, \text{mol} }$

Therefore, if we have 118.71 g of Sn, we will need 2 mol of HF to react with it.

Conclusion

In conclusion, we have answered several questions related to the chemical equation between tin (Sn) and hydrogen fluoride (HF). We have provided the balanced chemical equation, the molar masses of Sn and HF, and the equations for calculating the number of moles of Sn and HF required to react with each other. We have also provided example calculations to illustrate the process.