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The experiment calls for preparing one of two salts, either copper(ii) sulfate or potassium aluminum sulfate. The instructions for the copper(ii) sulfate preparation give a target of between 0.025 and 0.050 moles of the copper(ii) sulfate product. The instructions also specify to use an excess of sulfuric acid but not more than 5 times the amount required by the balanced equation. Sulfuric acid will be available as a 6.0 M H2SO4 solution. What is the volume of 6.0 M H2SO4 required by the balanced equation to react with the CuO to produce 0.025 moles of copper(ii) sulfate. Report the volume needed BEFORE allowing for excess in milliliters to two significant figures. Don not include units in the answer box. CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l)

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Answer:

4.2 mL

Explanation:

  • CuO(s) + H₂SO₄(aq) --> CuSO₄(aq) + H₂O(l)

We require 0.025 mol CuSO₄ and the ratio is 1:1, thus we need 0.025 mol H₂SO₄.

We know that the concentration of H₂SO₄ is 6.0 M, so we use the formula

  • Concentration = moles / Volume

And solve for volume:

  • Volume = moles / Concentration
  • Volume = 0.025 mol / 6.0 M
  • Volume = 4.2 x10⁻³ L = 4.2 mL

pstnonsonjoku

The volume of H2SO4 required is 0.0042L or 4.2 mL .

From the reaction equation;

CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l)

Number of moles of copper II oxide  = 0.025 moles

We can see that the reaction is 1:1 as such, the amount of H2SO4 reacted = 0.025 moles of H2SO4

We also know that;

Number of moles = concentration × volume

Hence;

volume = Number of moles/concentration

number of moles = 0.025 moles

concentration = 6.0 M

volume = 0.025 moles/6.0 M

Volume = 0.0042L or 4.2 mL

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