What mass of 95% pure CaCO3 will be require neutralise 50 mL of 0.5M HCl solution according to the following reaction?
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l) [Calculate upto second place of decimal point]

Option 1 - <p>1.25 g<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>1.32 g<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>3.65 g<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>9.50 g</p>
8 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
Correct Option - 2
Detailed Solution:

Moles of HCl = (50 / 1000) * 0.5 = 0.025 moles
So, moles of CaCO? used = 0.025 / 2 = 0.0125 moles = 1.25 g
95% (Total mass of CaCO? ) = 1.25 g
Total mass of CaCO? = 1.25 / 0.95 = 1.32 g

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