Considering the reaction stoichiometry, the amount of moles of NO that is produced when 27 moles of HNO₃ react is 6.75 moles.
<h3 /><h3>Reaction stoichiometry</h3>
The balanced reaction is:
3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Cu: 3 moles
- HNO₃: 8 moles
- Cu(NO₃)₂: 3 moles
- NO: 2 moles
- H₂O: 4 moles
<h3>Amount of moles of NO produced</h3>
It is possible to determine the the amount of moles of NO produced by a rule of three: if by stoichiometry 8 moles of HNO₃ produce 2 moles of NO, if 27 moles of HNO₃ react how many moles of NO will be formed?
<u><em>amount of moles of NO= 6.75 moles</em></u>
In summary, the amount of moles of NO that is produced when 27 moles of HNO₃ react is 6.75 moles.
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Answer:
28.11g
Explanation:
Step 1:
Background understanding:
From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10^23 atoms. This also indicates that 1 mole of helium (He) contains 6.02x10^23 atoms.
1 mole of He = 4g
Step 2:
Determination of the mass of He that contain 4.23x10^24 atoms. This is illustrated below:
4g of He contains 6.02x10^23 atoms.
Therefore Xg of He will contain 4.23x10^24 atoms i.e
Xg of He = (4x4.23x10^24)/6.02x10^23
Xg of He = 28.11g
Therefore, 28.11g of He contains 4.23x10^24 atoms
CaCl2(aq) + K2CO3(aq) → 2 KCl(aq) + CaCO3(aq)
1.12 g
2.23 g
0.896 g
4.47 g
1.12 g
Hope it helps :)
Answer:
Rise in volume = 72.7 mL
Explanation:
Given data:
Volume of water = 50.0 mL
Mass of brass piece = 193 g
Rise in volume = ?
Solution:
First of all we will calculate the volume of brass.
d = m/v
d = density
m = mass
v = volume
8.5 g/mL = 193 g/ v
v = 193 g/ 8.5 g/mL
v = 22.71 mL
Rise in volume = volume of water + volume of brass
Rise in volume = 50.0 mL + 22.7 mL
Rise in volume = 72.7 mL
