Question

(a) Calculate the mass of CaCO3 required to react completely with 25mL of
0.75 M HCl.
(b) Calculate volume of CO2 released at STP in this reaction.

Answer 1

Answer:

a)The1000mL of HCl=27.375g then the 1 ml of solution contains HCl

1000×1

27.375

25ml of solution which contains HCl=

1000×25

27.375

=0.684g

The chemical equation can be given as:

CaCO

3(s)

+2HCl→CaCl

2

+CO

2

+H

2

O

2mol of HCl reacts with 1 mol of CaCO

3

The amont of CaCO

3

reacted is given by,

71

100

×0.684g=0.9639g≈0.94g

b)Calculate volume of CO2 produced. The volume of one mole of CO2 produced is 24 dm^3 at room temperature and pressure. Alternatively, if your reaction took place at standard temperature and pressure (273 K, 1 atm), then the molar volume is 22.4 dm^3.

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Hope it helps you

Answer 2

Answer:

caco +nacl -sodium nirare = sodium nitrate