Question

The question is in the photo✨✨✨(Sorry) In the answer: Please leave all the numbers after the decimal point

Answer 1

36.2L of N2 can be produced.

1st) According to the stoichiometry of the reaction, 2 moles of NaN3 produce 3 moles of N2. Using the molar mass of NaN3 (65.0g/mol ) and N2 (28.0g/mol) we can convert the moles to mass, and we can see that with 130.0g of NaN3 we can produce 84.0g of N2.

Now, we can use a mathematical rule of three to calculate the grams of N2 that can be produced from 71.0g of NaN3:

`\begin{gathered} 130.0gNaN_3-84.0gN_2 \\ 71.0gNaN_3-x=(71.0gNaN_3\cdot84.0gN_2)/(130.0gNaN_3) \\ x=45.9gN_2 \end{gathered}`

So, 45.9g of N2 are produced from 71.0g of NaN3.

2nd) It is necessary to convert the grams of N2 produced to moles, so we can use it in the Ideal gas equation:

`\begin{gathered} 28.0gN_2-1mol \\ 45.9gN_2-x=(45.9gN_2\cdot1mol)/(28.0gN_2) \\ x=1.6\text{mol} \end{gathered}`

Now we know that 1.6mol of N2 are produced.

3rd) To calculate the volume of N2, it is necessary to use the Ideal gas equation and replace the values of Pressure (P), Temperature (T, in Kelvin) and Number of moles (n):

`\begin{gathered} P\mathrm{}V=n\mathrm{}R\mathrm{}T \\ 1.30\text{atm}\cdot V=1.6\text{mol}\cdot0.082(atm\cdot L)/(mol\cdot K)\cdot359K \\ V=\frac{1.6mol\cdot0.082(atm\cdot L)/(mol\cdot K)\cdot359K}{1.30\text{atm}} \\ V=36.2L \end{gathered}`

Finally, 36.2L of N2 can be produced from 71.0g of NaN3.