Question

Mg3N2+3H2O-3MgO+2NH3 If 17 g of magnesium nitride is used, what volume of ammonia gas would be collected at 20°C and 0.989 atm?​

Answer 1

Step-by-step explanation:

what are the unit given for the questions

Answer 2

The volume of ammonia gas that would be collected if 17 g of magnesium nitride is used, is 8.42 L

How to calculate the volume of ammonia gas collected?

Step 1: Calculate the mole of ammonia gas produced from the reaction. Details below:

• Mass of
  `Mg_3N_2` (m) = 17 g
• Molar mass of
  `Mg_3N_2` (M) = 98.2 g/mol
• Mole of
  `Mg_3N_2` = m / M = 17 / 98.2 = 0.173 mole
• Mole of ammonia gas,
  `NH_3` = ?

`Mg_3N_2\ +\ 3H_2O\ \rightarrow\ 3MgO\ +\ 2NH_3`

From the balanced equation above,

1 mole of
`Mg_3N_2` reacted to form 2 mole of
`NH_3`

Therefore,

0.173 mole of
`Mg_3N_2` will react to form = 0.173 × 2 = 0.346 mole of
`NH_3`

Step 2: Calculate the volume of ammonia gas,
`NH_3` collected. This is shown below:

• Temperature (T) = 20 °C = 20 + 273 = 293 K
• Pressure (P) = 0.989 atm
• Gas constant (R) = 0.0821 atm.L/molK
• Mole of ammonia gas,
  `NH_3` collected (n) = 0.346 mole
• Volume of ammonia gas,
  `NH_3` collected (V) =?

`V = (nRT)/(P) \\\\V = (0.346\ *\ 0.0821\ *\ 293)/(0.989) \\\\V = 8.42\ L`