The reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) to form silver carbonate (Ag2CO3) and sodium nitrate (NaNO3) is as shown in Equation 1. Roza weighed approximately 2.33 g Na2CO3 (106 g/mol)
Na2CO3(aq) + 2AgNO3(aq) Ag2CO3(s) + 2NaNO3(aq) (Equation 1)
(i) Calculate the mass of AgNO3 (molecular weight = 169.9g/mol) that is needed to convert 2.33 g of Na2CO3 (molecular weight =106 g/mol) to Ag2CO3.
(ii) Calculate the expected yield of Ag2CO3 in g.
The correct answer and explanation is :
(i) Calculation of the mass of AgNO₃ needed:
To determine how much AgNO₃ is needed to react with 2.33 g of Na₂CO₃, we first need to use stoichiometry. According to Equation 1:
[
\text{Na}_2\text{CO}_3 (aq) + 2\text{AgNO}_3 (aq) \rightarrow \text{Ag}_2\text{CO}_3 (s) + 2\text{NaNO}_3 (aq)
]
This equation tells us that 1 mole of Na₂CO₃ reacts with 2 moles of AgNO₃. Now, we will follow these steps:
- Find the moles of Na₂CO₃:
The molecular weight of Na₂CO₃ is given as 106 g/mol. To find the moles of Na₂CO₃ in 2.33 g, use the formula:
[
\text{moles of Na}_2\text{CO}_3 = \frac{\text{mass of Na}_2\text{CO}_3}{\text{molar mass of Na}_2\text{CO}_3}
]
[
\text{moles of Na}_2\text{CO}_3 = \frac{2.33\ \text{g}}{106\ \text{g/mol}} = 0.02197\ \text{mol}
]
- Use the stoichiometric ratio to find the moles of AgNO₃:
From the balanced equation, we know that 1 mole of Na₂CO₃ reacts with 2 moles of AgNO₃. Therefore:
[
\text{moles of AgNO}_3 = 2 \times \text{moles of Na}_2\text{CO}_3
]
[
\text{moles of AgNO}_3 = 2 \times 0.02197\ \text{mol} = 0.04394\ \text{mol}
]
- Calculate the mass of AgNO₃ needed:
Now, we can use the molecular weight of AgNO₃ (169.9 g/mol) to find the mass:
[
\text{mass of AgNO}_3 = \text{moles of AgNO}_3 \times \text{molar mass of AgNO}_3
]
[
\text{mass of AgNO}_3 = 0.04394\ \text{mol} \times 169.9\ \text{g/mol} = 7.47\ \text{g}
]
Thus, 7.47 g of AgNO₃ is needed to react with 2.33 g of Na₂CO₃.
(ii) Calculation of the expected yield of Ag₂CO₃:
To calculate the expected yield of Ag₂CO₃, we again use stoichiometry. From Equation 1:
[
\text{Na}_2\text{CO}_3 (aq) + 2\text{AgNO}_3 (aq) \rightarrow \text{Ag}_2\text{CO}_3 (s) + 2\text{NaNO}_3 (aq)
]
We already know that 1 mole of Na₂CO₃ produces 1 mole of Ag₂CO₃. Therefore, the moles of Ag₂CO₃ will be the same as the moles of Na₂CO₃.
- Determine the moles of Na₂CO₃:
From part (i), we calculated that the moles of Na₂CO₃ in 2.33 g are:
[
\text{moles of Na}_2\text{CO}_3 = 0.02197\ \text{mol}
]
- Find the moles of Ag₂CO₃ formed:
From the stoichiometric ratio, the moles of Ag₂CO₃ formed will be equal to the moles of Na₂CO₃ used:
[
\text{moles of Ag}_2\text{CO}_3 = 0.02197\ \text{mol}
]
- Calculate the mass of Ag₂CO₃ formed:
The molecular weight of Ag₂CO₃ is:
[
\text{Molar mass of Ag}_2\text{CO}_3 = 2(107.87) + 12.01 + 3(16.00) = 275.87\ \text{g/mol}
]
Now, we can calculate the expected mass of Ag₂CO₃:
[
\text{mass of Ag}_2\text{CO}_3 = \text{moles of Ag}_2\text{CO}_3 \times \text{molar mass of Ag}_2\text{CO}_3
]
[
\text{mass of Ag}_2\text{CO}_3 = 0.02197\ \text{mol} \times 275.87\ \text{g/mol} = 6.06\ \text{g}
]
Thus, the expected yield of Ag₂CO₃ is 6.06 g.
Explanation:
In this problem, we first calculated the amount of silver nitrate (AgNO₃) required to completely react with a given amount of sodium carbonate (Na₂CO₃). Using the stoichiometry from the balanced chemical equation, we found that for 2.33 g of Na₂CO₃, we needed 7.47 g of AgNO₃.
Next, we calculated the expected yield of silver carbonate (Ag₂CO₃) by considering the moles of Na₂CO₃ that would react to form Ag₂CO₃. Since the molar ratio between Na₂CO₃ and Ag₂CO₃ is 1:1, the moles of Ag₂CO₃ formed were the same as the moles of Na₂CO₃. Using the molar mass of Ag₂CO₃, we calculated that the expected yield was 6.06 g.
This process involves understanding stoichiometric relationships between reactants and products, and using the molecular weights and molar ratios to convert between grams, moles, and other units.