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Bingel [31]
3 years ago
13

In the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many liters of hydrogen gas will be produced from 75.0 milliliters o

f a 3.0 M HCl in an excess of Mg at STP?
a. 1.1 liters
b. 34 liters
c. 2.52 liters
d. 0.11 liters E. 0.23 liters
Chemistry
1 answer:
marta [7]3 years ago
6 0
The answer is <span>c. 2.52 liters.

</span>Molarity is a measure of the concentration of solute in a solution.
It can be expressed as moles of solute ÷ volume of solution:
c = n ÷V
where:
c - concentration of solute,
n - moles of solute
V - volume of solution
<span>
We have:
c = 3 M
V = 75 ml = 0.075 l
n = ?
Any gas at STP occupies 22.4 l/mol.</span><span>

If </span><span>c = n ÷V, then
n = c </span>× V
n = 3 M × 0.075 l
n = 0.225 moles HCl

In the reaction, there are 2 moles of HCl, so moles of H₂ would be the half of HCl moles:
n₁ = 0.255 moles HCl ÷ 2
n₁ = 0.1275 moles <span>H₂
</span>
Thus, if n₁ = 0.1275 moles H₂, and Mg<span> at STP occupies 22.4 l/mol</span><span>, then
</span>0.1275 moles H₂ × 22.4 l/mol = 2.52 l
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The empirical formula of the compound is = KBrO_3

The name of the compound is potassium bromate.

Explanation:

Mass of potassium = 4.628 g

Moles of potassium = \frac{4.628 g}{39 g/mol}=0.1187 mol

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\frac{0.3551 mol}{0.1182 mol}=3.0

The empirical formula of the compound is = KBrO_3

The name of the compound is potassium bromate.

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