Question

Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia reacts with O2 to give nitric oxide, NO.

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Then nitric oxide reacts with methane, CH4.

2NO(g) + 2CH4(g) → 2HCN(g) + 2H2O(g) + H2(g)

When 50.2 g of ammonia and 48.4 g of methane are used, how many grams of hydrogen cyanide can be produced? How many grams of which reactant remain at the end of both reactions? (You may assume that O2 is in excess in the first reaction.)

Answer 1

Answer:

The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

Explanation:

Given that,

Mass of ammonia = 50.2 g

Mass of methane = 48.4 g

Hydrogen cyanide, HCN, can be made by a two-step process

Ammonia reacts with O₂ to give nitric oxide NO.

The reaction is,

$4NH_{3}+5O_{2}\Rightarrow 4NO+6H_{2}O$

We need to calculate the mole of NO

Using given data,

2.25 g NH_{3}=\dfrac{50.2}{17}= 2.95\ mole\ NH_{3} [/tex]

$4\ mole NH_{3}\ glose 4\ mol NO$

2.95 mol NH₃ will produced 2.95 mol NO

Then nitric oxide reacts with methane,

The reaction is,

$2NO+2CH_{4}\Rightarrow 2HCN+2H_{2}O+H_{2}$

We need to calculate the mole of methane

Using given data,

$mole\ of\ methane=\dfrac{48.4}{16}$

$mole\ of\ methane = 3.03\ moles$

2 mole NO produced 2 mole HCN

2.95 mol NO will produced $\dfrac{2.95\times3.03}{3.03}$= 2.95 mol HCN

We need to calculate the mass of HCN

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times27$

$m= 79.65\ g$

The mass of HCN is 79.65 g.

We need to calculate the mass of NO

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times30$

$m= 88.5\ g$

Hence, The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.