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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?​

User Mikefrey
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2 Answers

10 votes

Step-by-step explanation:

what are the unit given for the questions

User Cliffordwh
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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

User Vladimir Baranov
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