> cao+co2 how many moles of cao form when 98.60g caco3 decompose? 98.60g caco3= mol cao

Hey I think it is going to be 0986 moles. Cuz we can see that no. of moles of CaCO3 which will decompose is equivalent to the no. of CaO . Now it's just the matter of finding the no. of moles of CaCO3 .

0.9852 moles of CaO

Explanation:

Reaction equation for the decomposition of CaCO₃:

CaCO₃ → CaO + CO₂

The question asks how many moles of CaO form when 98.60g of CaCO₃ decompose.

We can see from the reaction equation that for every mol of CaCO₃, one mol of CaO will be produced (molar ratio 1:1)

So first we need to calculate how many moles are the 98.60g of CaCO₃:

Molar Mass of CaCO₃ = molar mass Ca + molar mass C + 3 * molar mass O

                                     = 40.078 + 12.011 + 3 * 15.999 = 100.086 g/mol

Moles of CaCO₃ = mass CaCO₃ / molar mass CaCO₃

Moles of CaCO₃ = 98.60 g / 100.086 g/mol = 0.9852 moles CaCO₃

As we said before for every mol of CaCO₃, one mol of CaO is produced.

So the decomposition of 0.9852 moles of CaCO₃ will produce 0.9852 moles of CaO.

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