Question

How many liters of ammonia (NH3), at 3.2 atm and 23C, must be used to produce of 2.65 grams of calcium hydride (CaH2). 6 Ca(s) + 2 NH3 (g)--> 3 CaH2(s) + Ca3N2(g) *

Answer 1

The answer to your question is V = 0.32 L

Step-by-step explanation:

Data

Volume of NH₃ = ?

P = 3.2 atm

T = 23°C

mass of CaH₂ = 2.65 g

Balanced chemical reaction

6Ca + 2NH₃ ⇒ 3CaH₂ + Ca₃N₂

Process

1.- Convert the mass of CaH₂ to moles

-Calculate the molar mass of CaH₂

CaH₂ = 40 + 2 = 42 g

42 g ------------------ 1 mol

2.65 g -------------- x

x = (2.65 x 1)/42

x = 0.063 moles

2.- Calculate the moles of NH₃

2 moles of NH₃ --------------- 3 moles of CaH₂

x --------------- 0.063 moles

x = (0.063 x 2) / 3

x = 0.042 moles of NH₃

3.- Convert the °C to °K

Temperature = 23°C + 273

= 296°K

4.- Calculate the volume of NH₃

-Use the ideal gas law

PV = nRT

-Solve for V

V = nRT / P

-Substitution

V = (0.042)(0.082)(296) / 3.2

-Simplification

V = 1.019 / 3.2

-Result

V = 0.32 L