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3 years ago

Question 2 of 50

1 answer:

5 0

The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

Write the balanced chemical equation for the given reaction.
Convert the mass of calcium carbonate into moles.
Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

$CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}$

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

$mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}$

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

$mol \ CaO \ = 0.2498 \ mol$

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

$mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g$

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

$\boxed {mass \ CaO \ = 14 \ g}$

Learn More

Learn more about stoichiometry brainly.com/question/12979299
Learn more about mole conversion brainly.com/question/12972204
Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

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Read 2 more answers

For the following reaction, if 19.7 g of H2, is reacted with excess CO in the laboratory, and 144.5 g of CH3OH is produced, what

GuDViN [60]

Step 1

<em>The reaction involved:</em>

CO + 2 H2 → CH3OH (completed and balanced)

------------

Step 2

<em>Data provided:</em>

19.7 g H2 (the limiting reactant)

Excess reactant = CO

144.5 g CH3OH = actual yield

----

<em>Data needed:</em>

The molar masses of:

H2) 2.00 g/mol

CH3OH) 32.0 g/mol

-----------

Step 3

The theoretical yield:

By stoichiometry,

CO + 2 H2 → CH3OH (The molar rate between H2 and CH3OH = 2:1)

2 x 2.00 g H2 --------- 32.0 g CH3OH

19.7 g H2 --------- X

X = 19.7 g H2 x 32.0 g CH3OH/2 x 2.00 g H2

X = 157.6 g CH3OH (The theoretical yield)

-----------

Step 4

The % yield is defined as follows:

$\begin{gathered} \text{ \%yield = }\frac{Actual\text{ yield}}{Theoretical\text{ yield}}x100\text{ } \\ \text{ \%yield = }\frac{144.5\text{ g}}{157.6\text{ g}}x100\text{ = 91.7 \% = 92 \% approx.} \end{gathered}$

Answer: d. 93% (it is the nearest value in comparison to my result)

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