Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
Answer:
HCl + Ca(OH)2 = CaCl2 + H2O - Chemical Equation Balancer.
Answer: The coefficients for the given reaction species are 1, 6, 2, 3.
Explanation:
The given reaction equation is as follows.
Now, the two half-reactions can be written as follows.
Reduction half-reaction: 
This will be balanced as follows.
... (1)
Oxidation half-reaction: 
This will be balanced as follows.
... (2)
Adding both equation (1) and (2) we will get the resulting equation as follows.
Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.
