Answer: The mass of
dissolve can be, 7.25 grams.
Explanation : Given,
Mass of
= 5.30 g
Molar mass of
= 36.5 g/mol
Molar mass of
= 100 g/mol
First we have to calculate the moles of
.
![\text{Moles of }HCl=\frac{\text{Given mass }HCl}{\text{Molar mass }HCl}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DHCl%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DHCl%7D%7B%5Ctext%7BMolar%20mass%20%7DHCl%7D)
![\text{Moles of }HCl=\frac{5.30g}{36.5g/mol}=0.145mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DHCl%3D%5Cfrac%7B5.30g%7D%7B36.5g%2Fmol%7D%3D0.145mol)
Now we have to calculate the moles of ![CaCO_3](https://tex.z-dn.net/?f=CaCO_3)
The balanced chemical equation is:
![CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)](https://tex.z-dn.net/?f=CaCO_3%28s%29%2B2HCl%28aq%29%5Crightarrow%20CaCl_2%28aq%29%2BH_2O%28l%29%2BCO_2%28g%29)
From the reaction, we conclude that
As, 2 mole of
react with 1 mole of ![CaCO_3](https://tex.z-dn.net/?f=CaCO_3)
So, 0.145 mole of
react with
mole of ![CaCO_3](https://tex.z-dn.net/?f=CaCO_3)
Now we have to calculate the mass of ![CaCO_3](https://tex.z-dn.net/?f=CaCO_3)
![\text{ Mass of }CaCO_3=\text{ Moles of }CaCO_3\times \text{ Molar mass of }CaCO_3](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DCaCO_3%3D%5Ctext%7B%20Moles%20of%20%7DCaCO_3%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DCaCO_3)
![\text{ Mass of }CaCO_3=(0.0725moles)\times (100g/mole)=7.25g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DCaCO_3%3D%280.0725moles%29%5Ctimes%20%28100g%2Fmole%29%3D7.25g)
Therefore, the mass of
dissolve can be, 7.25 grams.