1. 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 65 mL of 0.15 M HCl?
2. What mass of calcium carbonate is needed for complete reaction with the HCl in (a)?
To find the number of moles of HCl in 65 mL of 0.15 M HCl, we can use the equation:
moles = concentration × volume
Given that the concentration of HCl is 0.15 M and the volume is 65 mL, we can substitute these values into the equation to find the number of moles of HCl:
moles = 0.15 M × 0.065 L
First, we need to convert the volume from milliliters (mL) to liters (L) by dividing by 1000:
moles = 0.15 M × (65/1000) L
moles = 0.15 M × 0.065 L
moles = 0.00975 mol
Therefore, there are 0.00975 moles of HCl in 65 mL of 0.15 M HCl.
Now, to calculate 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:
2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g)
We can see that the molar ratio between HCl and CaCO3 is 2:1. This means that for every 2 moles of HCl, we need 1 mole of CaCO3.
Since we have already determined that there are 0.00975 moles of HCl, we can use this value to calculate the moles of CaCO3 needed:
moles of CaCO3 = 0.00975 moles of HCl / 2
moles of CaCO3 = 0.004875 moles
To calculate the mass of calcium carbonate, we need to use its molar mass. The molar mass of calcium carbonate (CaCO3) is 100.0869 g/mol.
mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3
mass of CaCO3 = 0.004875 moles × 100.0869 g/mol
mass of CaCO3 = 0.4876 g
Therefore, approximately 0.4876 grams of calcium carbonate is needed for complete reaction with the HCl in 65 mL of 0.15 M HCl.
To solve these questions, we can use the given information and utilize some basic stoichiometry calculations.
1. Moles of HCl in 65 mL of 0.15 M HCl:
We'll use the equation:
Moles = Concentration (M) x Volume (L)
Given:
Concentration of HCl (C) = 0.15 M
Volume of HCl (V) = 65 mL = 65/1000 L (convert mL to L)
Moles of HCl = C x V
Moles of HCl = 0.15 M x (65/1000) L
Moles of HCl = 0.00975 moles
Therefore, there are 0.00975 moles of HCl in 65 mL of 0.15 M HCl.
2. Mass of calcium carbonate needed for complete reaction:
From the balanced equation, we see that the molar ratio between HCl and CaCO3 is 2:1. This means that for every 2 moles of HCl, we need 1 mole of CaCO3.
Since we know the number of moles of HCl from part 1 (0.00975 moles of HCl), we can calculate the moles of CaCO3 needed by dividing it by 2.
Moles of CaCO3 = Moles of HCl / 2
Moles of CaCO3 = 0.00975 moles / 2
Moles of CaCO3 = 0.004875 moles
Now, we need to convert the moles of CaCO3 to grams using its molar mass. The molar mass of CaCO3 is:
Molar mass of CaCO3 = (atomic mass of Ca) + (atomic mass of C) + (3 x atomic mass of O)
Atomic mass of Ca = 40.08 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.00 g/mol
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (3 x 16.00 g/mol)
Molar mass of CaCO3 = 100.09 g/mol
Now, we can calculate the mass of CaCO3:
Mass of CaCO3 = Moles of CaCO3 x Molar mass of CaCO3
Mass of CaCO3 = 0.004875 moles x 100.09 g/mol
Mass of CaCO3 = 0.487 g
Therefore, approximately 0.487 grams of calcium carbonate is needed for complete reaction with the HCl in part 1.
moles=.15M*.065L= ????
you need half that number of moles of CaCO3
massCaCO3=1/2 (.15*.065)*formulamassCaCO3