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amm1812
4 years ago
9

Consider separate samples of water and carbon dioxide, each with the same mass. Which contains the greater number of molecules?

Chemistry
1 answer:
Ahat [919]4 years ago
7 0

Answer: The sample that contains the greater number of molecules is water.

Explanation: To calculate the number of molecules, it is used the Avogadro's number ( 6.02*10^{23} particles/molecules). So, considering that the mass to water and carbon dioxide is 1g (it can be any other number), the relationship between moles and molar weight is:

moles =\frac{mass}{molar weight}

To water: moles = \frac{1g}{18 g/mol} = 0.05 moles.

Therefore,

1 mol ---- 6.02*10^{23} molecules

0.05 moles ---- x

x = 0.30*10^{23} molecules of water.

To carbon dioxide: moles = \frac{1g}{44 g/mol} = 0.02 moles.

Therefore,

1 mol ---- 6.02*10^{23} molecules

0.02 moles ---- y

y = 0.12*10^{23} molecules of carbon dioxide.

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During an experiment, 102 grams of calcium carbonate react with an excess amount of hydrochloric acid. If the percent yield of t
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96.31 g.

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<em>CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O,</em>

1.0 mole of CaCO₃ reacts with 2.0 moles of HCl to produce 1.0 mole of CaCl₂, 1.0 mole of CO₂, and 1.0 mole of H₂O.

  • We need to calculate the no. of moles of (104 g) of CaCO₃:

<em>no. of moles of CaCO₃ = mass/molar mass </em>= (102 g)/(100.08 g/mol) = <em>1.019 mol.</em>

<u><em>Using cross multiplication:</em></u>

1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.

∴ 1.019 mole of CaCO₃ produce → 1.019 mole of CaCl₂.

∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.019 mol)(110.98 g/mol) = 113.1 g.

∵ percent yield of the reaction = [(actual yield)/(theoretical yield)] x 100.

Percent yield of the reaction = 85.15%, theoretical yield = 113.1 g.

<em>∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100</em> = [(85.15%)(113.1 g)] / 100 = <em>96.31 g.</em>

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