What type of substance reacts with limestone (caco3) and produces carbon dioxide gas?
a) a neutral substance
b)a base
c)an acid
d)both an acid and a base

Option C= an acid

CaCO₃ is a very wide spread compound, and is found in chalk, lime, marble etc. This substance plays a very important role for human beings, as calcium carbonate is also widely used in the food industry, as a natural white colorant. CaCO₃ is also used for the manufacture of paper and plastic, in construction, and in many other spheres.

Calcium carbonate is a white substance (solid crystals) in powder or solid form, which is capable of reacting with water, but does not completely dissolve in it. But When any acid is reacted with calcium carbonate, the salt of the anionic part of the acid with calcium is formed. And H2CO3 is formed which may later break down to H2O+CO2, hence net reaction will be,

2HCl + CaCO3 → CaCl2 + H2CO3  

H2CO3 → H2O (l) + CO2 (g)

When calcium carbonate reacts with sulfuric acid, it forms water, carbon dioxide and calcium sulfate.

1. CaCO3 (s) + H2SO4 (l) → H2CO3 (aq) + CaSO4 (s)

2. H2CO3 (aq) → H2O (l) + CO2 (g)

So…

CaCO3 (s) + H2SO4 (l) → H2O (l) + CO2 (g) + CaSO4 (s)

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol