Question

Page
Add. How many particles of HCl are required to
react with 35 gm of CaCO3? ​

Answer 1

Step-by-step explanation:

From the chemical equation, we can see that for every mole of CaCO3, 2 moles of HCl react. This means that if you want them to react exactly in proportion, the number of moles of HCl needed is double that of CaCO3.

n(HCl)=2⋅n(CaCO3)

The molar mass (MM) of calcium carbonate is:

MM(CaCO3)=40.1+12.0+3⋅16.0=100.1 gmol

The number of moles of calcium carbonate can now be calculated:

n(CaCO3)=massMM=51.0100.1=0.509 mol

So, now we double the answer to get the amount of HCl needed:

n(HCl)=2⋅n(CaCO3)=2⋅0.509=1.02 mol

Rearranging the concentration formula we can get the volume needed in litres:

C=nV⇒V=nC

V=1.020.306=3.33 L

Convert to mL:

V=3.33⋅103 mL

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