32. A chemical reaction in which calcium carbonate CaCO3 is decomposed, results in the
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I found the attached image with the same statement of your question and think it may be very useful for you that I use it to show how to answer this question (furthermore I think it may be the same reaction that you forgot to include).
As you can see, there are one image on the left side and other image on hte right side of the figure.
Those images contains drawings that represent molecules and a legend that permit you to distinguish the kind of atoms in each molecule.
Using that, you can indicate the chemical reaction as the transformation of the molecules on the left side onto the molecules on the right side:
Left side:
3 molecules of CH4 and 3 molecules of N2Cl4
Right side:
3 molecules of CCl4, 3 molecules of N2 and 6 molecules of H2
That is represented as:
3CH4 + 3 N2Cl4 -----> 3 CCl4 + 3N2 + 6H2
And that is the balanced chemical equation for the reaction shown in the figured attached.
I hope this is useful for you..
As you can see, there are one image on the left side and other image on hte right side of the figure.
Those images contains drawings that represent molecules and a legend that permit you to distinguish the kind of atoms in each molecule.
Using that, you can indicate the chemical reaction as the transformation of the molecules on the left side onto the molecules on the right side:
Left side:
3 molecules of CH4 and 3 molecules of N2Cl4
Right side:
3 molecules of CCl4, 3 molecules of N2 and 6 molecules of H2
That is represented as:
3CH4 + 3 N2Cl4 -----> 3 CCl4 + 3N2 + 6H2
And that is the balanced chemical equation for the reaction shown in the figured attached.
I hope this is useful for you..
Answer : 87.60 g of Urea.
Explanation : We need to use the mole fraction formula here;
So, we have values and substituting them we get, 29.45 =
(31.8)
Hence, = 0.926 (this is the mole fraction of water)
Therefore, mole fraction of urea = 1 - 0.926 = 0.074,
moles of water present = 329 g / 18.0 mol = 18.27 moles,
Now, we have 0.074 moles of urea / 0.926 moles of water = x moles of urea / 18.27 moles of water,
Therefore, x = 1.46 moles of urea.
And now, Mass of urea = moles of urea X molar mass of urea;
= 1.46 X 60.0 = 87.60 g
Hence the mass of urea to be dissolved in 329 g of water will be 87.60 g
Explanation : We need to use the mole fraction formula here;
So, we have values and substituting them we get, 29.45 =
Hence,
Therefore, mole fraction of urea = 1 - 0.926 = 0.074,
moles of water present = 329 g / 18.0 mol = 18.27 moles,
Now, we have 0.074 moles of urea / 0.926 moles of water = x moles of urea / 18.27 moles of water,
Therefore, x = 1.46 moles of urea.
And now, Mass of urea = moles of urea X molar mass of urea;
= 1.46 X 60.0 = 87.60 g
Hence the mass of urea to be dissolved in 329 g of water will be 87.60 g