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tiny-mole [99]
3 years ago
10

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH4) by the following process: Hydrogen cyanide is used t

o 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?
Chemistry
1 answer:
Liula [17]3 years ago
5 0

<u>Answer:</u> The mass of HCN that could be made is 6.183 grams

<u>Explanation:</u>

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

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

  • <u>For ammonia:</u>

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

  • <u>For oxygen gas:</u>

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

  • <u>For methane:</u>

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

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