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topjm [15]
3 years ago
9

If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?

Chemistry
1 answer:
Hitman42 [59]3 years ago
4 0
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
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Answer:

The balanced chemical equation is Mg + 2HCl ⇒ MgCl2 + H2

The molarity of the hydrochloric acid solution is HCl 0.04 M and the pH = 1.4.

The volume of hydrogen gas produced by the reaction of 0.510 g of Mg will be 0.482 L.

Explanation:

First, for the balanced equation you have to consider the oxidation state of the elements to find subscripts. Then you can find the correct coeficients. Mg= +2, Cl = -1.

Mg + HCL ⇒ MgCl2 + H2

For the molarity of the solution you have to notice tha if 0.510 grams of Mg reacts with 0,5 L of hydroclhoric acid, and from the previous equation 1 mol of Mg reacts with 2 mol HCl.

The atomic mass of Mg = 24.31 grs/mol

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If 1 mol of Mg reacts with 2 mol HCl, then 0.02 mol of Mg will react with

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Then to calculate the pH we use the formula pH = - log [H+]

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