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Dahasolnce [82]
3 years ago
10

How many moles of CO2 must dissolve in excess water to produce 12 moles of H2CO3?

Chemistry
1 answer:
vodka [1.7K]3 years ago
3 0

Answer:

12 moles of CO₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CO₂ + H₂O —> H₂CO₃

From the balanced equation above,

1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.

Finally, we shall determine the number of moles of CO₂ that will dissolve in water to produce 12 moles of H₂CO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.

Therefore, 12 moles of CO₂ will also dissolve in water to produce 12 moles of H₂CO₃.

Thus, 12 moles of CO₂ is required.

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5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?
sveticcg [70]

By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's  Law</h3>

Given Data

  • number of moles of Ne = 5.1169 mol
  • Pressure = 0.9148 atm
  • Temperature = 911 K

We know that the relationship between pressure and temperature is given as

PV = nRT

R = 0.08206

Making the volume subject of formula we have

V= nRT/P

Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

V = 382.522353554/0.9148

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Learn more about Boyle's law here:

brainly.com/question/469270

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2 years ago
If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce
Natali5045456 [20]
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)

=0.0282669621 g of O2 left over
5 0
3 years ago
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