Question

when 0.25 mg\\(_{(s)}\\) is added to 250ml of 3.00 mol/l hcl\\(_{(aq)}\\), hydrogen gas is produced. the magnesium becomes mg\\(^{2+}_{(aq)}\\) ions and stays in the solution. start by writing ionic and net ionic equations for the reaction. what is the of h\\(^+\\) ion h\\(^+) mol/l question 12 (1 point) retake question if 50 ml of h\\(_2\\)so\\(_4\\) yields 0.300 g of baso\\(_4\\) when reacts with excess of bacl\\(_2\\), what is the concentration of h\\(_2\\)so\\(_4\\) in mol/l bacl\\(_{2(aq)}\\) + h\\(_2\\)so\\(_{4(aq)}\\) ---> baso\\(_{4(s)}\\) + 2 hcl\\(_{(aq)}\\) mol/l

Explanation:

Step1: Calculate moles of BaSO₄

Molar mass of BaSO₄: $Ba$ (137.33) + $S$ (32.07) + 4×$O$ (16.00) = 137.33 + 32.07 + 64.00 = 233.4 g/mol.
Moles of BaSO₄ = $\frac{0.300\ g}{233.4\ g/mol}$ ≈ 0.001285 mol.

Step2: Relate moles of H₂SO₄ to BaSO₄

From the reaction: $BaCl_2 + H_2SO_4
ightarrow BaSO_4 + 2HCl$, the mole ratio of $H_2SO_4$ to $BaSO_4$ is 1:1. So moles of $H_2SO_4$ = 0.001285 mol.

Step3: Calculate concentration of H₂SO₄

Volume of H₂SO₄ = 50 mL = 0.050 L.
Concentration (mol/L) = $\frac{moles}{volume}$ = $\frac{0.001285\ mol}{0.050\ L}$ ≈ 0.0257 mol/L.

Answer:

0.0257 (or approximately 0.026)