What mass of calcium chloride in grams would be enough to produce `14*35` g of AgCl? (At. mass: Ca=40, Ag=108)

Options

A. `5*55 g`
B. `8*295 g`
C. `16*59 g`
D. `11*19g`

Explanation

To solve the problem of how much mass of calcium chloride (CaCl₂) is needed to produce 14.35 grams of silver chloride (AgCl), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between calcium chloride and silver nitrate can be represented as:
\[ \text{CaCl}_2 + 2 \text{AgNO}_3 \rightarrow \text{Ca(NO}_3)_2 + 2 \text{AgCl} \]

### Step 2: Calculate the molar mass of AgCl
The molar mass of AgCl can be calculated using the atomic masses:
- Atomic mass of Ag = 108 g/mol
- Atomic mass of Cl = 35.5 g/mol

Thus, the molar mass of AgCl is:
\[ \text{Molar mass of AgCl} = 108 + 35.5 = 143.5 \text{ g/mol} \]

### Step 3: Calculate the number of moles of AgCl produced
To find the number of moles of AgCl produced from 14.35 grams, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles of AgCl} = \frac{14.35 \text{ g}}{143.5 \text{ g/mol}} \approx 0.1 \text{ moles} \]

### Step 4: Use stoichiometry to find moles of CaCl₂ needed
From the balanced equation, we see that 1 mole of CaCl₂ produces 2 moles of AgCl. Therefore, the moles of CaCl₂ required can be calculated as follows:
\[ \text{Moles of CaCl}_2 = \frac{0.1 \text{ moles of AgCl}}{2} = 0.05 \text{ moles of CaCl}_2 \]

### Step 5: Calculate the molar mass of CaCl₂
Next, we calculate the molar mass of calcium chloride:
- Atomic mass of Ca = 40 g/mol
- Atomic mass of Cl = 35.5 g/mol (and there are 2 Cl atoms)

Thus, the molar mass of CaCl₂ is:
\[ \text{Molar mass of CaCl}_2 = 40 + 2 \times 35.5 = 111 \text{ g/mol} \]

### Step 6: Calculate the mass of CaCl₂ needed
Now, we can find the mass of CaCl₂ required using the number of moles and its molar mass:
\[ \text{Mass of CaCl}_2 = \text{moles} \times \text{molar mass} \]
\[ \text{Mass of CaCl}_2 = 0.05 \text{ moles} \times 111 \text{ g/mol} = 5.55 \text{ g} \]

### Final Answer
The mass of calcium chloride required to produce 14.35 grams of AgCl is **5.55 grams**.

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