What mass of calcium chloride in grams would be enough to produce 14.35 gm of AgCl? 5.55 gm 8.295 gm 16.5 gm 11.19 gm

To answer this question, we need to use stoichiometry, which is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

The balanced chemical equation for the reaction between calcium chloride (CaCl2) and silver nitrate (AgNO3) to produce silver chloride (AgCl) and calcium nitrate (Ca(NO3)2) is:

CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2

From the balanced equation, we can see that one mole of CaCl2 produces two moles of AgCl.

First, we need to find the molar mass of AgCl. The atomic mass of silver (Ag) is approximately 107.87 g/mol and that of chlorine (Cl) is approximately 35.45 g/mol. So, the molar mass of AgCl is 107.87 g/mol + 35.45 g/mol = 143.32 g/mol.

Next, we need to find out how many moles of AgCl are in 14.35 g. We can do this by dividing the mass of AgCl by its molar mass:

14.35 g ÷ 143.32 g/mol = 0.1 mol

Since the balanced equation tells us that one mole of CaCl2 produces two moles of AgCl, we need half as many moles of CaCl2 to produce 0.1 mol of AgCl, which is 0.05 mol.

Finally, we need to find the mass of 0.05 mol of CaCl2. The molar mass of CaCl2 is approximately 110.98 g/mol (40.08 g/mol for Ca and 35.45 g/mol for each Cl). So, the mass of CaCl2 needed is:

0.05 mol × 110.98 g/mol = 5.55 g

So, 5.55 g of calcium chloride would be enough to produce 14.35 g of AgCl.