In the reaction: <span>caco3(s) → cao(s) + co2(g), it is evident that
1 mol (which is 100 g) of CaCO3 yields 1 mol (which is 44 g) of CO2
Now, the molarity of CaCO3 present in the reaction system is
</span>=

=

= 0.45 mol
Thus, 0.45 mol of CaCO3 leads to the formation of 0.45 mol of CO2.
According to the ideal gas equation, we have PV = nRT
V =

.
Considering P = 645 torr = 0.8487 atm (because 1 atm = 760 torr)
In that case, V =

= 34.8 l
Answer:
3.816 × 10⁻³ M
Explanation:
A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?
We can derive the following relations:
- The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
- Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.
The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

The molarity of Cu²⁺ is:
