i want to neutralize a 10 ml 3 molar sulfuric acid using calcium carbonate but don't know how to do the math the explanation confused me.
(also i have to use sig fig and uncertainty)
The question posted under Joe says calcium bicarbonate and that's what I answered. For CaCO3 is is a different equation but the process is exactly the same.
Step 1. Write and balance the equation.
CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2
Step 2. How many mols do you need to neutralize; i.e., how mols is in 10 mL of 3M H2SO4.
mols = M x L. You know M and you know L. Solve for mols H2SO4.
Step 3. Using the coefficients in the balanced equation, convert mols H2SO4 to mols CaCO3.
?mols H2SO4 x (1 mol CaCO3/1 mol H2SO4) = ?mols H2SO4 x 1/1 so mols H2SO4 = mols CaCO3.
Step 4. Now convert mols CaCO3 to grams CaCO3.
g CaCO3 = mols CaCO3 x molar mass CaCO3.
This four-step procedure will work almost any stoichiometry problem in the book. I didn't write anything different here than I did with your Ca(HCO3)2 post, except of course I changed to CaCO3. If you don't understand a step or the reasoning behind the step explain what you don't understand and I can help you through that. This is as simple as it gets.
To neutralize a 10 mL solution of 3 M sulfuric acid (H2SO4) using calcium carbonate (CaCO3), you need to determine the amount of calcium carbonate required based on the stoichiometry of the reaction.
First, let's write the balanced chemical equation for the reaction:
H2SO4(aq) + CaCO3(s) -> CaSO4(s) + CO2(g) + H2O(l)
From the balanced equation, we can see that one mole of sulfuric acid reacts with one mole of calcium carbonate.
To calculate the amount of calcium carbonate needed, we need to convert the volume of sulfuric acid (in mL) to moles using the provided concentration of 3 M. Here's how:
1. Convert the volume from mL to L: 10 mL = 0.01 L
2. Calculate the number of moles: moles = volume (L) * concentration (M) = 0.01 L * 3 M = 0.03 moles of sulfuric acid (H2SO4)
Since the stoichiometry is 1:1, we need 0.03 moles of calcium carbonate (CaCO3) to neutralize the 0.03 moles of sulfuric acid.
Now, we need to calculate the mass of calcium carbonate using molar mass and apply significant figures and uncertainty.
1. Find the molar mass of calcium carbonate:
Ca: 1 atom * 40.08 g/mol = 40.08 g/mol
C: 1 atom * 12.01 g/mol = 12.01 g/mol
O3: 3 atoms * 16.00 g/mol = 48.00 g/mol
Total molar mass of CaCO3: 40.08 g/mol + 12.01 g/mol + 48.00 g/mol = 100.09 g/mol
2. Calculate the mass of calcium carbonate:
mass = moles * molar mass = 0.03 moles * 100.09 g/mol = 3.0027 g
To apply significant figures and uncertainty, you need to consider the measurement and uncertainties in the provided data. If the 10 mL volume measurement has an uncertainty of ±0.1 mL, you need to propagate the uncertainty throughout the calculations. However, since the uncertainty for concentration is not given, we will assume the value is exact.
Using the given volume uncertainty of ±0.1 mL:
1. Convert the volume uncertainty to L: ±0.1 mL = ±0.0001 L
2. Calculate the moles of sulfuric acid with the uncertainty:
moles = volume (L) * concentration (M)
Maximum moles: (0.01 L + 0.0001 L) * 3 M = 0.0303 moles
Minimum moles: (0.01 L - 0.0001 L) * 3 M = 0.0297 moles
3. Calculate the mass of calcium carbonate with the maximum and minimum moles:
Maximum mass: 0.0303 moles * 100.09 g/mol = 3.0303 g
Minimum mass: 0.0297 moles * 100.09 g/mol = 2.9703 g
Therefore, the mass of calcium carbonate required to neutralize the 10 mL 3 M sulfuric acid solution is 3.0027 g (±0.03 g).
Remember to use appropriate sig figs and round the final answer to match the level of uncertainty provided in the data.