Calcium carbonate reacts with HCl according to the following equation:
2HCl(aq)+CaCO3(s)→CaCl2(aq)+H2O(l)+CO2(g)
How many moles of HCl are in 58 mL of 0.13 M HCl?
What mass of calcium carbonate is needed for complete reaction with the HCl in (a)?
molesHCl=.058*.13
you need mole of calcium carbonate 1/2 the moles of HCl
that is NOT the correct answer for the mass of calcium carbonate needed for complete reaction with the HCl in part (a) I tried the same thing and a few other ways and all are wrong---any additional suggestions
hbl, I think you jumped the gun. mols CaCO3 = 1/2 mols HCl as Bob P wrote. He assumed you would be able to convert mols to grams.
Then mass CaCO3 = mols CaCO3 x molar mass CaCO3 = ?
Deez nutz 069
To find the number of moles of HCl in 58 mL of 0.13 M HCl, you can use the formula:
moles = volume (in liters) x concentration
Step 1: Convert the volume from milliliters (mL) to liters (L)
58 mL ÷ 1000 = 0.058 L
Step 2: Multiply the volume by the concentration to find the number of moles
moles = 0.058 L x 0.13 mol/L
moles = 0.00754 moles
Therefore, there are approximately 0.00754 moles of HCl in 58 mL of 0.13 M HCl.
To find the mass of calcium carbonate needed for complete reaction with the HCl, we need to use the stoichiometry of the balanced equation.
From the equation: 2 moles of HCl react with 1 mole of CaCO3.
Step 1: Convert the moles of HCl to moles of CaCO3 using the stoichiometry.
moles of CaCO3 = (moles of HCl) ÷ 2
moles of CaCO3 = 0.00754 moles ÷ 2
moles of CaCO3 = 0.00377 moles
Step 2: Calculate the molar mass of calcium carbonate (CaCO3).
Ca: 40.08 grams/mol
C: 12.01 grams/mol
O: 16.00 grams/mol (3 O atoms per molecule)
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (16.00 g/mol x 3)
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + 48.00 g/mol
Molar mass of CaCO3 = 100.09 g/mol
Step 3: Calculate the mass of calcium carbonate needed.
mass (g) = moles x molar mass
mass (g) = 0.00377 moles x 100.09 g/mol
mass (g) = 0.377 g
Therefore, approximately 0.377 grams of calcium carbonate is needed for complete reaction with the HCl in 58 mL of 0.13 M HCl.