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Answer :
Answer: 91.6
Explanation:
To determine the quantity of silver chloride formed, we need to balance the chemical equation for the reaction between calcium chloride and silver nitrate:`CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2`From the balanced equation above, we can see that 2 moles of silver chloride are produced for every 1 mole of calcium chloride. The molar mass of calcium chloride is 110.98 g/mol while that of silver chloride is 143.32 g/mol.To calculate the quantity of silver chloride formed, we can use the following steps:1. Calculate the number of moles of calcium chloride: ``` moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2 ``` ``` moles of CaCl2 = 35.5 g / 110.98 g/mol = 0.319 moles ``` 2. Calculate the number of moles of silver chloride using stoichiometry: ``` moles of AgCl = (moles of CaCl2) x (2 moles of AgCl / 1 mole of CaCl2) ``` ``` moles of AgCl = 0.319 x 2 = 0.638 moles ``` 3. Calculate the mass of silver chloride formed: ``` mass of AgCl = moles of AgCl x molar mass of AgCl ``` ``` mass of AgCl = 0.638 moles x 143.32 g/mol = 91.6 g ``` Therefore, 91.6 grams of silver chloride will form when 35.5 grams of calcium chloride reacts.
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Rewritten by : Barada
To determine the quantity of silver chloride formed, we need to balance the chemical equation for the reaction between calcium chloride and silver nitrate:
`CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2`
From the balanced equation above, we can see that 2 moles of silver chloride are produced for every 1 mole of calcium chloride. The molar mass of calcium chloride is 110.98 g/mol while that of silver chloride is 143.32 g/mol.
To calculate the quantity of silver chloride formed, we can use the following steps:
1. Calculate the number of moles of calcium chloride:
```
moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2
```
```
moles of CaCl2 = 35.5 g / 110.98 g/mol = 0.319 moles
```
2. Calculate the number of moles of silver chloride using stoichiometry:
```
moles of AgCl = (moles of CaCl2) x (2 moles of AgCl / 1 mole of CaCl2)
```
```
moles of AgCl = 0.319 x 2 = 0.638 moles
```
3. Calculate the mass of silver chloride formed:
```
mass of AgCl = moles of AgCl x molar mass of AgCl
```
```
mass of AgCl = 0.638 moles x 143.32 g/mol = 91.6 g
```
Therefore, 91.6 grams of silver chloride will form when 35.5 grams of calcium chloride reacts.
