Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia is reacted with O2 to give nitric oxide, NO.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Then nitric oxide is reacted with methane, CH4.
2 NO(g) + 2 CH4(g) → 2 HCN(g) + 2 H2O(g) + H2(g)
When 25.6 g of ammonia and 25.1 g of methane are used, how many grams of hydrogen cyanide can be produced?
Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia is reacted with O2 to give nitric oxide, NO.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
I would use this equation with simple stoichiometry to convert 25.6 g NH3 to grams NO. Here is a site that gives an example of a simple stoichiometry problem which I have posted.http://www.jiskha.com/science/chemistry/stoichiometry.html
Then nitric oxide is reacted with methane, CH4.
2 NO(g) + 2 CH4(g) → 2 HCN(g) + 2 H2O(g) + H2(g)
Then use grams NO and 25.1 g CH4 in a limiting reagent problem to determine the amount of HCN formed.
When 25.6 g of ammonia and 25.1 g of methane are used, how many grams of hydrogen cyanide can be produced?
I got the grams of NO, but how do you do the limiting reagent problem?
To find out the number of grams of hydrogen cyanide (HCN) produced, we need to determine the limiting reactant and then use stoichiometry to calculate the amount of HCN formed.
Let's start by finding out the limiting reactant. The limiting reactant is the reactant that is completely consumed and restricts the amount of product that can be formed. To determine the limiting reactant, we compare the amount of reactants used with their respective molar ratios.
First, we convert the given masses of ammonia (NH3) and methane (CH4) to moles using their molar masses:
Molar mass of NH3 = 17.03 g/mol
Moles of NH3 = 25.6 g / 17.03 g/mol ≈ 1.505 mol
Molar mass of CH4 = 16.04 g/mol
Moles of CH4 = 25.1 g / 16.04 g/mol ≈ 1.564 mol
Next, we compare the mole ratios of NH3 and CH4 from the balanced chemical equation:
From the equation:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
2 NO(g) + 2 CH4(g) → 2 HCN(g) + 2 H2O(g) + H2(g)
For every 4 moles of NH3, 2 moles of HCN are produced.
Moles of HCN produced from 1.505 mol NH3 = (1.505 mol NH3) x (2 mol HCN / 4 mol NH3) = 0.7525 mol HCN
For every 2 moles of CH4, 2 moles of HCN are produced.
Moles of HCN produced from 1.564 mol CH4 = (1.564 mol CH4) x (2 mol HCN / 2 mol CH4) = 1.564 mol HCN
Comparing the moles of HCN produced from NH3 and CH4, we can see that the limiting reactant is NH3 since it produces the fewer number of moles of HCN (0.7525 mol).
Now, using stoichiometry, we can calculate the mass of HCN produced. We'll convert the moles of HCN to grams using its molar mass:
Molar mass of HCN = 27.03 g/mol
Mass of HCN produced = 0.7525 mol HCN x 27.03 g/mol ≈ 20.355 g
Therefore, approximately 20.355 grams of hydrogen cyanide (HCN) can be produced when 25.6 g of ammonia (NH3) and 25.1 g of methane (CH4) are used in the given reaction.