7 would be correct because I divide it
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:
Barium
Explanation:
-Atomic #56
-5th element inn group 2
- soft
- silvery alkaline earth metal
Answer:
u already know the answer
Explanation:answer
Answer:
10 L of CO₂.
Explanation:
The balanced equation for the reaction is given below:
2CO + O₂ —> 2CO₂
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Finally, we shall determine the volume of CO₂ produced by the reaction of 10 L CO. This can be obtained as follow:
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Therefore, 10 L of CO will also react to produce 10 L of CO₂.
Thus, 10 L of CO₂ were obtained from the reaction.
