Question

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH4) by the following process: Hydrogen cyanide is used to prepare sodium cyanide, which is used in part to obtain gold from gold-containing rock. If a reaction vessel contains 5.90 g NH3, 11.0 g O2, and 4.67 g CH4, what is the maximum mass in grams of hydrogen cyanide that could be made, assuming the reaction goes to completion as written?

Answer 1

Answer: The mass of HCN that could be made is 6.183 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For ammonia:

Given mass of ammonia = 5.90 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{5.90g}{17g/mol}=0.347mol$

• For oxygen gas:

Given mass of oxygen gas = 11.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{11.0g}{32g/mol}=0.344mol$

• For methane:

Given mass of methane = 4.67 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{4.67g}{16g/mol}=0.292mol$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The mole ratio of the reactants are:

$NH_3:O_2:CH_4=2:3:2::1:1.5:1$

As, the moles of oxygen gas is the lowest. So, it is considered as the limiting reagent

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of HCN

So, 0.344 moles of oxygen gas will produce = $\frac{2}{3}\times 0.344=0.229mol$ of HCN

Now, calculating the mass of HCN from equation 1, we get:

Molar mass of HCN = 27 g/mol

Moles of HCN = 0.229 moles

Putting values in equation 1, we get:

$0.229mol=\frac{\text{Mass of HCN}}{27g/mol}\\\\\text{Mass of HCN}=(0.229mol\times 27g/mol)=6.183g$

Hence, the mass of HCN that could be made is 6.183 grams