please check if I did this right ...
calculate the mass of white solid CaCO3 that forms when 98ml of a 0.1M Ca(NO3)2 solution is mixed with 54 ml of a 2.5M Na2CO3 solution.
Ca(NO3)2 + 2NaCO3 -> Ca(CO3)2 + 2NaNO3
.098*0.1 =.0090
.054*2.5 =.135
limiting reactant is Ca(NO3)2
.0090 * 160.1 = 1.44g
the balanced equation should be this:
Ca(NO3)2+ Na2CO3>>CaCO3 + 2NaNO3
You had two formulas wrong.
Thank you
so with those corrections my new calculations would be
.0090*100.05 = .90045g
Hopefully this time it is right.
Obviously I am not chemisrty gifted.
so with those corrections my new calculations would be
.0090*100.05 = .90045g
The 0.0090 number you have comes from 0.098 x 0.1 = 0.0098 and you made a typo when you entered it. Then 0.0098 x 100.05 = 0.98 g
i have a question of why did you put 2 after the parentheses Ca(CO3)2
isn't Ca has the charge number of 2+
and the CO3 has the charge of negative 2
To calculate the mass of white solid CaCO3 that forms when 98 mL of a 0.1 M Ca(NO3)2 solution is mixed with 54 mL of a 2.5 M Na2CO3 solution, you first need to determine the limiting reactant. The limiting reactant is the one that will be completely consumed in the reaction, thereby determining the maximum amount of product that can be formed.
To find the limiting reactant, you can use the stoichiometry of the balanced equation provided. In this case:
Ca(NO3)2 + 2Na2CO3 -> CaCO3 + 4NaNO3
To compare the amounts of each reactant, you can use the concept of molarity. Molarity is calculated by dividing the moles of solute by the volume of the solution in liters.
For Ca(NO3)2:
Molarity = 0.1 M
Volume = 98 mL = 0.098 L
Number of moles of Ca(NO3)2 = Molarity * Volume
= 0.1 M * 0.098 L
= 0.0098 moles of Ca(NO3)2
For Na2CO3:
Molarity = 2.5 M
Volume = 54 mL = 0.054 L
Number of moles of Na2CO3 = Molarity * Volume
= 2.5 M * 0.054 L
= 0.135 moles of Na2CO3
Since the stoichiometry of the balanced equation indicates that the ratio of Ca(NO3)2 to Na2CO3 is 1:2, we compare the number of moles of Ca(NO3)2 to twice the number of moles of Na2CO3.
Moles of Ca(NO3)2 = 0.0098 moles
Moles of Na2CO3 = 2 * 0.135 moles = 0.27 moles
Since the moles of Ca(NO3)2 (0.0098 moles) are less than the moles of Na2CO3 (0.27 moles), Ca(NO3)2 is the limiting reactant.
Now, to find the mass of CaCO3 that forms, you can use the molar mass of CaCO3, which is 100.1 g/mol.
Mass of CaCO3 = Moles of Ca(NO3)2 * Molar mass of CaCO3
= 0.0098 moles * 100.1 g/mol
= 0.98 g
Therefore, the mass of white solid CaCO3 that forms when the given solutions are mixed is 0.98 grams.