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Nat2105 [25]
3 years ago
14

How many milliliters of a 1.25 molar hydrochloric acid (HCl) solution would be needed to react completely with 60.0 grams of cal

cium metal? (2 points)
Ca (s) + 2HCl (aq)yields CaCl2 (aq) + H2 (g)
Chemistry
1 answer:
stich3 [128]3 years ago
4 0

<u>Answer:</u>

2400 mL

<u>Explanation:</u>

Ca + 2HCl \implies CaCl_2 + H_2

According to this equation, the stoichiometric ratio between Ca and HCl for the complete reaction is 1:2.

We know that the number of moles of Ca can be calculated using the mole formula. (<em>number of moles = mass / molar mass</em>)

Moles of Calcium = \frac{60}{40} = 1.5 mol

So the moles of HCl = 1.5 \times 2 = 3.0 mol

<em>Volume of HCl solution = Moles of HCl/ concentration of HCl</em>

Volume of HCl solution = \frac{3}{1.25} = 2400 mL

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The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

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