The relative molecular mass of calcium carbonate (CaCO3) is 100. What is the minimum volume of 2.0 M hydrochloric acid (HCl) that would be needed to completely react with 2.0 g of calcium carbonate?

CaCO3 + 2HCl > CaCl3 + CO2 + H2O

2grams/100= .02 moles
so you need .04 moles HCl

molesHCl=volume*Molarity
volume = .04/2= .02 liters or 20 ml

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Well, let's calculate that, shall we? The first step is to convert the mass of calcium carbonate to moles. So, we divide the mass (2.0 g) by the relative molecular mass (100 g/mol). And voila! We get 0.02 moles of CaCO3.

Now, since the reaction between calcium carbonate and hydrochloric acid is 1:2, we know that we need twice as many moles of HCl as CaCO3. Therefore, we need 0.04 moles of HCl.

Next, we can use the formula Molarity = Moles / Volume (in liters) to find the volume of the hydrochloric acid needed. We rearrange the formula to Volume = Moles / Molarity.

Plugging in the numbers, we get Volume = 0.04 moles / 2.0 M = 0.02 L.

So, the minimum volume of 2.0 M hydrochloric acid needed to react with 2.0 g of calcium carbonate is 0.02 liters. Now, that's just a rough estimate. In reality, you might need slightly more because reactions, like people, can be a little unpredictable sometimes.

To find the minimum volume of hydrochloric acid (HCl) needed to react with calcium carbonate (CaCO3), we can use the concept of stoichiometry and the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is:

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

From the balanced equation, you can see that one mole of calcium carbonate reacts with 2 moles of hydrochloric acid.

Step 1: Calculate the number of moles of calcium carbonate (CaCO3) using its relative molecular mass.

The relative molecular mass of calcium carbonate (CaCO3) is given as 100 g/mol. Therefore,

Number of moles of CaCO3 = Mass of CaCO3 / Relative molecular mass of CaCO3
= 2.0 g / 100 g/mol
= 0.02 mol

Step 2: Determine the number of moles of hydrochloric acid (HCl) required using the stoichiometry of the balanced equation.

Since the stoichiometric ratio is 1:2 between CaCO3 and HCl, we need twice the number of moles of HCl.

Number of moles of HCl = 2 * Number of moles of CaCO3
= 2 * 0.02 mol
= 0.04 mol

Step 3: Calculate the volume of 2.0 M hydrochloric acid (HCl) needed using its molarity and the equation:

Volume (in liters) = Number of moles / Molarity

Volume of HCl = 0.04 mol / 2.0 M
= 0.02 L
= 20 mL

Therefore, the minimum volume of 2.0 M hydrochloric acid (HCl) needed to completely react with 2.0 g of calcium carbonate is 20 mL.

Thanks! Bob