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jarptica [38.1K]
3 years ago
7

Aspartame (c14h18n2o5) is a solid used as an artificial sweetener. its combustion produces carbon dioxide gas, liquid water, and

nitrogen gas. what is the coefficient of nitrogen gas in the balanced equation for the reaction? (balance the equation with the smallest possible whole number coefficients.)
Chemistry
1 answer:
Law Incorporation [45]3 years ago
5 0

Aspartame (C₁₄H₁₈N₂O₅) is a solid used as an artificial sweetener. its combustion produces carbon dioxide gas, liquid water, and nitrogen gas

C₁₄H₁₈N₂O₅ + 16O₂-----> 14CO₂ + 9H₂O + N₂.

As it can be seen from the equation, that the coefficient of nitrogen gas in the balanced equation for the reaction is 1.

So the answer here is 1 only that is coefficient of nitrogen gas in the balanced equation for the reaction is 1.

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Do electrons have psrticle nature or wave nature?​
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2 years ago
If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃  = 18.1g

Mass of Cu₂O  = 90.4g

Unknown:

Limiting reactant  = ?

Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

  Number of moles = \frac{mass}{molar mass}  

Molar mass of Cu₂O = 2(63.6) + 16  = 143.2g/mol

Molar mass of NH₃  = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = \frac{18.1}{143.2}   = 0.13moles

Number of moles of NH₃   = \frac{90.4}{17}   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

    0.13 mole of Cu₂O  will produce 0.13 mole of N₂

    Mass  = 0.13 x (2 x 14) = 3.64g

5 0
3 years ago
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