The answer is B; I recently completed the test on Edge <3
Although multiple values are given, our focus is on HCl.
<span>We have 215 mL (0.215 L) of 0.300 M HCl fully consumed in the reaction. It's important to recall that the number of moles is found by multiplying volume by molarity:</span>
moles = 0.215 L × 0.300 M
<span>moles = 0.0645 moles of HCl</span>
Answer:
The original halide's formula is SrCl₂.
Explanation:
- The chemistry reaction's balanced equation is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.
- Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.
- The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.
- n = mass / molar mass, thus n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ is calculated using mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- Calculating the atomic mass of halide X, we find = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This identifies the atomic mass of Cl.
- Consequently, the original halide's formula is SrCl₂.
