If 10.3 g of Mg3N2 is treated with water, what volume of gas would be collected at STP

Question

Chemistry

Cordell King

13 March, 20:48

If 10.3 g of Mg3N2 is treated with water, what volume of gas would be collected at STP

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Answers (2)

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Sincere Jefferson · Answer 1

Mg3N2 + 6H2O → 3Mg (OH) 2 + 2 NH3

relative weight of Mg3N2 = (3x24) + (2x14) = 100 g/mole

mole of Mg3N2 = 10,3 g/100 g/mole = 0,103 mole

mole of NH3 (g) = 2/1 x 0,103 mole = 0,206 mole

volume NH3 (g) which collected at STP = mole NH3 (g) x 22,4 litre/mole = 0,206 mole x 22,4 litre/mole = 4,6144 litre.

Jackson · Answer 2

Use the ideal gas law: PV=nRt To find volume, move to V = (nRt) / P N=# of moles (to find, 10.3 g/100.929g) =.102 moles R=ideal gas constant (this value is 0.082056 (Liters*atmospheres) / (Kelvin*mol) T = temp which at stp is 273 Kelvin P = 1 atmosphere V = ((.102 moles) (0.082056 Latm/Kmol) (273 K)) / 1 atm