How many milliliters of 0.480 M HCl are needed to react with 54.8 g of CaCO3?...

Question

Question

How many milliliters of 0.480 M HCl are needed to react with 54.8 g of CaCO3?

2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)

mL?

Answer 1

Answer #1

nHcl=2. n(caco3)

The molar mass of caco3=100.1g/mol

Moles of caco3=54.8g/100.1g/mol=0.54mol

So now we double the answer to get the amount of Hcl needed

n HCl=2 n(caco3)=2*0.54=1.1mol

Concentration formula

C=n/V

V=n/C

V=1.1/0.480

V=2.28L

V=2280ML

Answer 2

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