Question

Hydrogen gas and nitrogen gas come together to form ammonia. If 25.0 g of nitrogen is mixed with hydrogen, what volume of hydrogen is needed to complete the reaction? (grams to volume)

Answer 1

60.0L of hydrogen are needed

Step-by-step explanation:

Based on the reaction:

3H₂ + N₂ ⇄ 2NH₃

3 moles of hydrogen react 1 mole of nitrogen.

To solve this question we have to find the moles of nitrogen. With the moles of nitrogen we can find the moles of hydrogen. Using PV = nRT at STP conditions we can find the volume as follows:

Moles Nitrogen -Molar mass: 28g/mol-

25.0g N₂ * (1mol / 28g) = 0.893 moles N₂

Moles hydrogen:

0.893 moles N₂ * (3mol H₂ / 1mol N₂) = 2.679 moles H₂

Volume hydrogen:

PV = nRT

V = nRT / P

Where V is volume in liters,

n are moles of the gas: 2.679 moles

R is gas constant: 0.082atmL/molK

T is absolute temperature: 273.15K at STP

P is 1atm at STP

Replacing:

V = 2.679mol*0.082atmL/molK*273.15K / 1atm

V =

60.0L of hydrogen are needed