Approximately 9.41 grams of CaCO3 can be formed from the reaction of 10.00 grams of NaCO3.
To determine the amount of CaCO3 formed, we first need to find the limiting reactant in the reaction between NaCO3 and CaCl2. The balanced chemical equation is:
\[ NaCO_3 + CaCl_2 \rightarrow CaCO_3 + 2NaCl \]
First, find the moles of each reactant:
1. **NaCO3:**
\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{moles} = \frac{10.00 \, \text{g}}{105.99 \, \text{g/mol}} \approx 0.094 \, \text{mol} \]
2. **CaCl2:** Since CaCl2 is in excess, it will not limit the reaction.
Now, examine the stoichiometry of the reaction. The molar ratio between NaCO3 and CaCO3 is 1:1. This means that for every 1 mole of NaCO3, 1 mole of CaCO3 is formed.
The moles of CaCO3 formed will be the same as the moles of NaCO3 reacted since they have a 1:1 ratio.
\[ \text{moles of CaCO3 formed} = 0.094 \, \text{mol} \]
Now, find the mass of CaCO3 formed:
\[ \text{mass} = \text{moles} \times \text{molar mass} \]
\[ \text{mass} = 0.094 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 9.41 \, \text{g} \]
Therefore, approximately 9.41 grams of CaCO3 can be formed from the reaction of 10.00 grams of NaCO3.
The probable question may be:
How many grams of CaCO3 can be formed from the reaction of 10.00g of NaCO3 with an excess of CaCl2 according to the following reaction? NaCO3 + CaCl2 ---> CaCO3 + 2NaCl