What volume of 1.34 mol/L hydrochloric acid, HCI, is expected to react completely with 6.89 g of calcium?

Ca + 2HCl --> CaCl2 + H2

mols Ca = grams/atomic mass
mols HCl = twice that since 1 mol Ca = 2 mols HCl
Then M HCl = mols HCl/L HCl and this can be arranged to L HCl = mols HCl/M HCl which let's you solve for L HCl.

To find the volume of 1.34 mol/L hydrochloric acid (HCI) that will react completely with 6.89 g of calcium, we need to use stoichiometry and molar ratios.

First, let's determine the molar mass of calcium (Ca). You can find this information from the periodic table. The molar mass of calcium is 40.08 g/mol.

Next, we need to convert the mass of calcium (6.89 g) to moles. We can use the formula:

moles = mass / molar mass

moles of calcium = 6.89 g / 40.08 g/mol ≈ 0.172 mol

According to the balanced chemical equation:

2 HCI + Ca → CaCl2 + H2

we can see that 1 mole of calcium reacts with 2 moles of hydrochloric acid. This means that the moles of hydrochloric acid required to react completely with 0.172 mol of calcium are 2 * 0.172 mol = 0.344 mol.

Now, we can calculate the volume of hydrochloric acid needed. We have a concentration of 1.34 mol/L, which means that for every 1 L of hydrochloric acid, there are 1.34 moles.

volume (L) = moles / concentration (mol/L)
volume (L) = 0.344 mol / 1.34 mol/L ≈ 0.256 L

Therefore, the volume of 1.34 mol/L hydrochloric acid required to react completely with 6.89 g of calcium is approximately 0.256 liters.