Question

How many milliliters of
`H_(2)` gas at STP are required to fully hydrogenate 1.68 g of
`C_(6) H_(8)N_(2)` (adiponitrile) according to the following hydrogenation reaction scheme?


`C_(6)H_(8)N_(2)(l) + 4H_(2)(g)` →
`C_(6) H_(16)N_(2)(s)`

Answer 1

To fully hydrogenate 1.68 g of adiponitrile, 1.88 liters of H2 gas at STP are required.

Step-by-step explanation:

To find the volume of hydrogen gas (H2) required to fully hydrogenate 1.68 g of adiponitrile (C6H8N2), we need to use stoichiometry and the ideal gas law. In the given reaction, it is mentioned that 1 mole of adiponitrile requires 4 moles of hydrogen gas, so we can use this ratio to calculate the moles of hydrogen gas required.

First, we calculate the moles of adiponitrile:

1.68 g C6H8N2 × (1 mol C6H8N2 / 80.14 g C6H8N2) = 0.02098 mol C6H8N2

Using the stoichiometry, we find the moles of hydrogen gas:

0.02098 mol C6H8N2 × (4 mol H2 / 1 mol C6H8N2) = 0.08392 mol H2

Now we can use the ideal gas law to find the volume of hydrogen gas at STP (Standard Temperature and Pressure):

0.08392 mol H2 × (22.4 L / 1 mol H2) = 1.88 L

Therefore, 1.88 liters of H2 gas at STP are required to fully hydrogenate 1.68 g of adiponitrile.

Answer 2

The answer is

21.5 L

.

So, start with the balanced chemical equation for the decomposition of hydrochloric acid

2

HCl

→

H

2

+

Cl

2

Notice that you have a

2:1

mole ratio between

HCl

and

Cl

2

, which means that every 2 moles of the former will produce 1 mole of the latter. The number of moles of

HCl

you have is.

Step-by-step explanation:

I don't know if that is but I try my best just correct me if I'm wrong thank you!!