Question

Calculate the volume (mL) of 1.0 M metal nitrate solution required to prepare 3.0 g of CaCO₃. Again, using the balanced chemical equation you should be able to relate the moles of pigment to the moles of metal nitrate and from that calculate the volume of 1.0 M metal nitrate solution required.

Balanced equation: Ca(NO₃)₂ (aq) + K₂CO₃ (aq) ↔ CaCO₃ (s) + 2KNO₃ (aq)

Answer 1

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass of CaCO3 = 3.0 g

we have below equation to be used:

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(3.0 g)/(100.09 g/mol)

= 3.0*10^-2 mol

Balanced equation is:

Ca(NO3)2 (aq) + K2CO3 (aq) ↔ CaCO3 (s) + 2KNO3 (aq)

from above reaction,

moles of Ca(NO3)2 required = moles of CaCO3 formed

= 3.0*10^-2 mol

now use:

Molarity * volume = moles of Ca(NO3)2

1.0 M * volume = 3.0*10^-2 mol

volume = 3.0*10^-2 L

volume = 30 mL

Answer: 30 mL