Step-by-step explanation:
The given data is as follows.
m = 19.2 g, I = 15 A
The given reaction equation will be as follows.
![CaCl_(2) \rightarrow Ca^(2+) + 2Cl^(-)](https://img.qammunity.org/2020/formulas/chemistry/college/dt8cak830ke4j1j47kc21c5of0ihpe3ktu.png)
![Ca(s) \rightarrow Ca^(2+) + 2e^(-)](https://img.qammunity.org/2020/formulas/chemistry/college/xgee43l7ikt9rs7uovvao0jgjqgtejsf0y.png)
As, 1 mole will give
of charge. So, calculate the amount of charge deposited by 19.2 g as follows.
![2 * 96500 C * (19.6 g)/(40.07 g/mol)](https://img.qammunity.org/2020/formulas/chemistry/college/b5ne26f1ytg9pvdg1hvnfnwd8gjzdm49av.png)
= 94404.791 C
Hence, 94404.791 C are needed to produce 19.2 g of Ca metal.
Therefore for 15 A, minutes required to produce 94404.791 C and 19.2 g of Ca(s) will be calculated as follows.
=
![94404.791 C * (1 Amp sec)/(1 C) * (1)/(15 Amp) * (1 min)/(60 sec)](https://img.qammunity.org/2020/formulas/chemistry/college/f6p4y8z7p1h7hr3broyg4sxgd18oe079jz.png)
= 104.9 mins
Thus, we can conclude that 104.9 minutes are required if a cell runs at 15 A, how many minutes will it take to produce 19.2 g of Ca(s).