To find the area of something you have to multiply the length x the with x the height. Or if it gives to the base then you multiply the base x the height.
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Given:
Moles of H2 = 0.300
Moles of I2 = 0.400
Moles of HI = 0.200
Keq = 870
To determine:
Amounts of the mixture at equilibrium
Explanation:
H2(g) + I2(g) ↔ 2HI(g)
Initial 0.3 0.4 0.2
Change -x -x +2x
Eq (0.3-x) (0.4-x) (0.2+2x)
Keq = [HI]²/[H2][I2]
870 = (0.2+2x)²/(0.3-x)(0.4-x)
x = 0.29 moles
Amounts at equilibrium:
[HI] = 0.2 + 2(0.29) = 0.78 moles
[H2] = 0.3-0.29 = 0.01 moles
[I2] = 0.4-0.29 = 0.11 moles
Answer:
C2= 0.16M
Explanation:
C1= 2M, V1= 20ml, C2= ?, V2= 250ml
Applying dilution formula
C1V1= C2V2
2×20 =C2×250
C2= 0.16M
Taking into account the reaction stoichiometry, 2.405 moles of NaCl are formed when 127.49 grams of Na₂CO₂ reacts with excess of FeCl₃.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
2 FeCl+ 3 Na₂CO₂ → Fe₂(CO₃)₃ + 6 NaCl
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:
- FeCl: 2 moles
- Na₂CO₂: 3 moles
- Fe₂(CO₃)₃: 1 mole
- NaCl: 6 moles
The molar mass of the compounds is:
- FeCl: 91.3 g/mole
- Na₂CO₂: 106 g/mole
- Fe₂(CO₃)₃: 291.7 g/mole
- NaCl: 58.45 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- FeCl: 2 moles ×91.3 g/mole= 182.6 grams
- Na₂CO₂: 3 moles ×106 g/mole= 318 grams
- Fe₂(CO₃)₃: 1 mole ×291.7 g/mole= 291.7 grams
- NaCl: 6 moles ×58.45 g/mole= 350.7 grams
<h3>Mass of NaCl produced</h3>
The following rule of three can be applied: if by reaction stoichiometry 318 grams of Na₂CO₂ form 6 moles of NaCl, 127.49 grams of Na₂CO₂ form how many moles of NaCl?
<u><em>moles of NaCl= 2.405 moles</em></u>
Then, 2.405 moles of NaCl are formed when 127.49 grams of Na₂CO₂ reacts with excess of FeCl₃.
Learn more about the reaction stoichiometry:
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