CHEMISTRY | MOLE CONCEPT | Question 08
A 2.00g sample 0f ammonia is mixed with 4.00g of Oxygen.
Which is the limiting reactant and how much excess reactant remain after the reaction has stopped?
Balanced equation for the reaction:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(s)
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DATA GIVEN:
Mass of NH3(g) = 2.00g
Mass of O2(g) = 4.00g
Asked
• To find ions which is limiting
• The remain amount of excess reactant
Equation
NH3(g) + 5O2 4NO(g) + 6H2O(s)
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To find ions which is limiting
To convert 2.00g into mole
Mole = mass /Molar mass of NH3
= 2.00g /17g/mole
= 0.117647 mole
Mole ratio = 0.125 mole /5 mole
= 0.025
The limiting is O2, the excess (NH3) remain from mole equivalent between NH3 and O2
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The remain amount of excess reactant
4 moles of NH3 Ξ 5 mole of O2
?? Ξ 0.125 mole of O2
= 4 moles × 0.12 mole /5 mole
= 0.1 mole of NH3
To convert 0.1 mole of NH3 into mass
Mass = mole × molar mass
= 0.1 mole × 17g/mole
= 1.7g reacted of NH3
Mass remain = original mass- reacted mass
= 2.00g – 1.7g
= 0.3g
0.3g of ammonia (NH3) remain unreacted.
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