How many milliliters of 0.258 M NiCl2 solution are needed to react completely with 20.0 mL of 0.153 M Na2CO3 solution? How many grams of NiCO3 will be formed? The reaction is
Na2CO3(aq) +NiCl2 (aq)----> NiCO3(s)+2NaCl
I solved for the 1st part, 11.8 M NiCl2
I am not sure where to begin to solve for grams.
What's 11.8 M NiCl2? The problem says M NiCl2 is 0.258. You must mean 11.8 mL.
mols Na2CO3 = M x L = 0.153 x 0.020 = ?
mols NiCO3 formed = same since there is 1 mol NiCO3 for 1 mol Na2CO3.
grams NiCO3 = mols NiCO3 x molar mass NiCO3.
But note that I get 11.86 mL which I would round to 11.9 mL to three significant figures.
To solve for the number of milliliters of 0.258 M NiCl2 solution needed to react completely with 20.0 mL of 0.153 M Na2CO3 solution, you can use the concept of stoichiometry.
1. Start by writing down the balanced chemical equation for the reaction:
Na2CO3(aq) + NiCl2(aq) -> NiCO3(s) + 2NaCl(aq)
2. Determine the stoichiometric ratio between Na2CO3 and NiCl2 from the balanced equation. In this case, it is 1:1. This means that for every 1 mole of Na2CO3, you need 1 mole of NiCl2 for complete reaction.
3. Calculate the number of moles of Na2CO3 in the 20.0 mL of 0.153 M solution:
moles Na2CO3 = volume (L) x concentration (M)
moles Na2CO3 = 0.020 L x 0.153 M = 0.00306 mol Na2CO3
4. Since the stoichiometric ratio between Na2CO3 and NiCl2 is 1:1, you need an equal number of moles of NiCl2 for complete reaction:
moles NiCl2 = 0.00306 mol Na2CO3
5. Use the concentration of the NiCl2 solution to calculate the volume of the 0.258 M NiCl2 solution needed:
volume NiCl2 (L) = moles NiCl2 / concentration NiCl2
volume NiCl2 = 0.00306 mol / 0.258 M = 0.0119 L = 11.9 mL
Therefore, 11.9 milliliters of 0.258 M NiCl2 solution are needed to completely react with 20.0 mL of 0.153 M Na2CO3 solution.
To find the grams of NiCO3 formed:
6. Note that the stoichiometric ratio between NiCl2 and NiCO3 is 1:1. This means that for every 1 mole of NiCl2, 1 mole of NiCO3 is formed.
7. Calculate the number of moles of NiCO3 formed using the moles of NiCl2 calculated earlier:
moles NiCO3 = moles NiCl2 = 0.00306 mol
8. Calculate the molar mass of NiCO3:
Ni: 58.69 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
molar mass NiCO3 = 58.69 + 12.01 + (3 × 16.00) = 118.69 g/mol
9. Finally, calculate the mass of NiCO3 formed:
mass NiCO3 = moles NiCO3 × molar mass NiCO3
mass NiCO3 = 0.00306 mol × 118.69 g/mol = 0.363 g
Therefore, 0.363 grams of NiCO3 will be formed in the reaction.
To solve for the number of milliliters of 0.258 M NiCl2 solution needed to react completely with 20.0 mL of 0.153 M Na2CO3 solution, you need to use the concept of stoichiometry.
First, you need to determine the balanced chemical equation for the reaction:
Na2CO3(aq) + NiCl2(aq) → NiCO3(s) + 2NaCl(aq)
From the equation, you can see that the stoichiometric ratio between Na2CO3 and NiCl2 is 1:1. This means that for every 1 mole of Na2CO3, you need 1 mole of NiCl2 to react completely.
Next, you need to calculate the number of moles of Na2CO3 present in the 20.0 mL of 0.153 M Na2CO3 solution. To do this, use the formula:
moles of solute = concentration (M) × volume (L)
moles of Na2CO3 = 0.153 M × 0.020 L = 0.00306 moles
Since the stoichiometric ratio is 1:1, the number of moles of NiCl2 required is also 0.00306 moles.
Now, you can calculate the volume of 0.258 M NiCl2 solution needed by using the formula:
volume (L) = moles of solute / concentration (M)
volume of NiCl2 = 0.00306 moles / 0.258 M ≈ 0.0119 L
To convert the volume to milliliters, multiply it by 1000:
volume of NiCl2 = 0.0119 L × 1000 = 11.9 mL
Therefore, approximately 11.9 milliliters of 0.258 M NiCl2 solution are needed to react completely with 20.0 mL of 0.153 M Na2CO3 solution.
To calculate the number of grams of NiCO3 formed, you need to use the molar mass of NiCO3, which is the sum of the atomic masses of Ni, C, and 3 times the atomic mass of O.
Ni: 58.69 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of NiCO3 = (58.69 g/mol) + (12.01 g/mol) + 3 × (16.00 g/mol) = 74.69 g/mol
Since the stoichiometric ratio between NiCl2 and NiCO3 is 1:1, the number of moles of NiCO3 formed is also 0.00306 moles.
To calculate the mass of NiCO3, you can use the formula:
mass = moles × molar mass
mass of NiCO3 = 0.00306 moles × 74.69 g/mol = 0.227 grams
Therefore, approximately 0.227 grams of NiCO3 will be formed when 20.0 mL of 0.153 M Na2CO3 solution reacts completely with 11.9 mL of 0.258 M NiCl2 solution.