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posledela
3 years ago
12

A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )

was back titrated with 23.60 mL of 0.0980 M NaOH ( aq ) . How many moles of HCl react with the carbonate?
Chemistry
1 answer:
Alex73 [517]3 years ago
3 0

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = \frac{0.1200}{1000}\times 50.0 moles = 0.00600 moles

Neutralization reaction (back titration): NaOH+HCl\rightarrow NaCl+H_{2}O

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = \frac{0.0980}{1000}\times 23.60 moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

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General Formulas and Concepts:

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<u>Stoichiometry</u>

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Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

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

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<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 1.8 \ mol \ Mg(\frac{6.022 \cdot 10^{23} \ atoms \ Mg}{1 \ mol \ Mg})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

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