Question

From the following balanced equation, 4 NH3(g) +50,(9) how many grams of O, had to react to form 3.78 g H,O? 4 NO(g) + 6H,O(1) Select the correct answer below: O 2.568 O 4.545 O O 5.596 08.06 FEEDBACK

Answer 1

Solution:-

4NH₃ (g) + 5O₂ (g)   $\rightarrow$ 4NO (g) + 6H₂O (l)

It is clear from the balanced chemical equation, that 5 moles of O₂ had to react to form 6 moles of H₂O. Or, we can say that, to produce 1 mole of H​​​​​​₂O , the number of moles of O₂ which had to react = 5/6 moles

Number of moles in 3.78 g of H₂O = Mass ÷ molar mass = (3.78 g) ÷ (18.01 g/mole) = 0.209 moles.

(Note:- molar mass of H₂O = 18.01 g/mole)

Now, the number of moles of O₂ which had to react to produce 0.209 mole of H₂O is = (5/6) × 0.209 moles = 0.175 moles

Mass of 0.175 mole of O₂ = Number of moles × molar mass = (0.175 mole) × (32 g/mol) = 5.59 g

(Note:- molar mass of H₂​​​​​​​O = 32 g/mole)

Hence, 5.59 g of O₂ had to react to form 3.78 g of H₂O

"Option iii) : 5.59 g" is correct answer.

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