Question

I need help with part b the answer for part a is 15

Answer 1

Answer: 27g of NaCN would be necessary to produce the lethal dose of HCN

Step-by-step explanation:

The question requires us to calculate the mass of NaCN necessary to produce enough HCN for a lethal dose, considering that this lethal dose corresponds to 15g of HCN and the following balanced chemical equation:

`2NaCN_((s))+H_2SO_(4(aq))\rightarrow Na_2SO_(4(aq))+2HCN_((g))`

The problem presented can be seen as a mass to mass stoichiometry problem, thus we'll need to follow these steps to solve it:

1) determine the molar mass of NaCN and HCN;

2) from the mass of HCN given, calculate the correspondent number of moles of HCN;

3) from the number of moles of HCN calculated, determine the required number of moles of NaCN;

4) convert the obtained number of moles of NaCN to its correspondent mass.

1) Calculating the molar mass of NaCN and HCN:

We can obtain the molar masses from the atomic masses of Na, C, N and H, which are 22.99, 12.01, 14.01 and 1.007 amu, respectively.

The molar masses can be calculated as:

`\begin{gathered} molar\text{ mass \lparen NaCN\rparen = 1}*22.99+1*12.01+1*14.01=49.01g/mol \\ \\ molar\text{ mass \lparen HCN\rparen = 1}*1.007+1*12.01+1*14.01=27.03g/mol \end{gathered}`

Therefore, the molar mass of NaCN and HCN are 49.01 and 27.03 g/mol, respectively.

2) Calculating the number of moles of HCN

Considering that the lethal dose corresponds to 15g of HCN, we can calculate the number of moles necessary for this lethal dose from the molar mass of HCN:

27.03g HCN -------------------- 1 mol HCN

15g HCN ------------------------- x

Solving for x, we'll have:

`x=15g\text{ HCN}*\frac{1mol\text{ HCN}}{27.03g\text{ HCN}}=0.55mol\text{ HCN}`

Therefore, 0.55 moles of HCN need to be used for a lethal dose.

3) Calculating the number of moles of NaCN:

From the balanced chemical equation given, we can see that 2 moles of HCN are produced when 2 moles of NaCN react, thus we can calculate how many moles of NaCN would be necessary to produce 0.55 moles of HCN:

2 mol HCN ---------------------- 2 mol NaCN

0.55 mol HCN ----------------- y

Solving for y, we'll have:

`y=0.55mol\text{ HCN}*\frac{2mol\text{ NaCN}}{2mol\text{ HCN}}=0.55mol\text{ NaCN}`

Therefore, 0.55 moles of NaCN would be necessary to produce the lethal dose of HCN.

4) Calculating the mass of NaCN:

Given that we need 0.55 moles of NaCN and the molar mass of this compound is 49.01 g/mol, we can calculate the required mass of NaCN:

1 mol NaCN -------------------- 49.01g NaCN

0.55 mol NaCN --------------- z

Solving for z, we'll have:

`z=0.55mol\text{ NaCN}*\frac{49.01g\text{ NaCN}}{1mol\text{ NaCN}}=26.9g\text{ NaCN}\cong27g\text{ NaCN}`

Therefore, 27g of NaCN would be necessary to produce the lethal dose of HCN.