The question is incomplete. The complete question is :
Hydrogen 
gas and oxygen 
gas react to form water vapor 
. Suppose you have 11.0 mol of 
and 13.0 mol of 
in a reactor. Calculate the largest amount of 
that could be produced. Round your answer to the nearest 0.1 mol .
Solution :
The balanced reaction for reaction is :
11.0 13.0
11/2 13/1 (dividing by the co-efficient)
6.5 mol 13 mol (minimum is limiting reagent as it is completely consumed during the reaction)
Therefore, is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for will be produced = number of moles of
= 11.0 mol
Answer:
b. 0.22 L
Explanation:
Step 1: Write the balanced equation
4 HNO₃(l) ⇒ 4 NO₂(g) + 2 H₂O(l) + O₂(g)
Step 2: Calculate the moles corresponding to 2.5 g of HNO₃
The molar mass of HNO₃ is 63.01 g/mol.
2.5 g × 1 mol/63.01 g = 0.040 mol
Step 3: Calculate the moles of O₂ produced from 0.040 moles of HNO₃
The molar ratio of HNO₃ to O₂ is 4:1. The moles of O₂ produced are 1/4 × 0.040 mol = 0.010 mol.
Step 4: Calculate the volume corresponding to 0.010 moles of O₂
Assuming the reaction takes place at standard temperature and pressure, the volume of 1 mole of O₂ is 22.4 L.
0.010 mol × 22.4L/1 mol = 0.22 L
Answer:
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Explanation:
The answer to your question is FALSE
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