Solution Found!
How many milliliters of 0.383 M HCI are needed to react
Chapter 3, Problem 74P(choose chapter or problem)
How many milliliters of \(0.383 M\) HCl are needed to react with \(16.2 g\) of \(\mathrm{CaCO}_{3}\)?
\(2 \mathrm{HCl}(\mathrm{aq})+\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
Equation Transcription:
CaCO3
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
Text Transcription:
0.383 M
16.2 g
CaCO_3
2HCl(aq) + CaCO_3(s) rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)
Questions & Answers
QUESTION:
How many milliliters of \(0.383 M\) HCl are needed to react with \(16.2 g\) of \(\mathrm{CaCO}_{3}\)?
\(2 \mathrm{HCl}(\mathrm{aq})+\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
Equation Transcription:
CaCO3
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
Text Transcription:
0.383 M
16.2 g
CaCO_3
2HCl(aq) + CaCO_3(s) rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)
ANSWER:
Solution 74P
Here, we have to calculate the millilitres of 0.383 M HCl are needed to react with 16.2 g of CaCO3.
Step 1 of 2
The given chemical reaction
2HCl (aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O (l).
First let’s convert the grams of CaCO3 into moles.
Given
Mass of CaCO3 = 16.2 gm.
Molar mass of CaCO3= 100.09 g
Moles of CaCO3 = Given mass of CaCO3
= 16.2 gm
= 0.161854 mol of CaCO3
Therefore, moles of CaCO3 is 0.161854 mol.
Let’s convert the moles of CaCO3 into moles of HCl
Moles of HCl = moles of CaCO3
= 0.161854 mol of CaCO3
= 0.323708 mol of HCl
Therefore, moles of HCl is 0.323708 mol.
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