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2 years ago

How many moles of calcium, Ca, are in 5.00 g of calcium?

Chemistry

1 answer:

Rom4ik [11]2 years ago

3 0

<h3>Answer:</h3>

0.125 mol Ca

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

5.00 g Ca

<u>Step 2: Identify Conversions</u>

Molar mass of Ca - 40.08 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 5.00 \ g \ Ca(\frac{1 \ mol \ Ca}{40.08 \ g \ Ca})$
Multiply: $\displaystyle 0.12475 \ mol \ Ca$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.12475 mol Ca ≈ 0.125 mol Ca

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2 years ago

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Gala2k [10]

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Alright the very first thing you need to do is balance the equation:

2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O

Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.

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Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.

Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.

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Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio) and the volume we just calculated for NaCl.

(0.75M) x (0.094L) = C2 x (0.329L)

Rearrange equation to solve for C2:

<u>(0.75M) x (0.094L)</u> = C2

(0.329L)

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<u>When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.</u>

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