Question

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: 2NH3 1g2 1 3O2 1g2 1 2CH4 1g2 h 2HCN1g2 1 6H2O1g2 If 5.00 3 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?

Answer 1

Answer: 1687.19 g of $HCN$ and 1124,69 g of $H_2O$

Explanation:

We have the following reaction:

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

To use it, we must convert mass to moles of reactants, by diving each mass between the molar mass of the substance:

$n_{NH_3} = \frac{mass\ of\ NH_3}{molar\ mass\ of\ NH_3} = \frac{5000g}{17.031 \frac{g}{mol} }=293.58\ mol\\n_{O_2} = \frac{mass\ of\ O_2}{molar\ mass\ of\ O_2} = \frac{3000g}{31.999 \frac{g}{mol} }=93.65\ mol\\\\n_{CH_4} = \frac{mass\ of\ CH_4}{molar\ mass\ of\ CH_4} = \frac{103000g}{16.04 \frac{g}{mol} }=6421.45\ mol$

Now we must identify the limiting reactant, the one that will be totally consumed in the reaction. We take the one with fewer moles and calculate how much moles of the other ones are needed to consume it all, using a rule of three:

$3\ moles\ of O_2 - 2\ moles\ of NH_3 - 2\ moles\ of\ CH_4\\93.65\ moles\ of O_2 - 62.43\ moles\ of NH_3 - 62.43\ moles\ of\ CH_4$

So there is enough $NH_3$ and $CH_4$ to consume the oxygen. In case there wasn´t enough of one of the other reactants, that one would be the limiting one.

Now, we can use the limiting reactant to calculate the moles of products obtained, again using a rule of three:

$3\ moles\ of O_2 - 2\ moles\ of HCN - 2\ moles\ of\ H_2O\\93.65\ moles\ of O_2 - 62.43\ moles\ of HCN - 62.43\ moles\ of\ H_2O$

Lastly, we can convert moles to mass by multipling by the molar mass of each substance:

$m_{HCN} = moles\ of\ HCN * molar\ mass\ of\ HCN = 62.43\ moles * 27.0253 \frac{g}{mol} =1687.19\ g\\m_{H_2O} = moles\ of\ H_2O * molar\ mass\ of\ H_2O = 62.43\ moles * 18.01528 \frac{g}{mol} =1124,69\ g$