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lozanna [386]
3 years ago
11

A chemist prepares hydrogen fluoride by means of the following reaction:

Chemistry
1 answer:
natita [175]3 years ago
4 0

Answer:

39.3%

Explanation:

CaF2 + H2SO4 --> CaSO4 + 2HF

We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:

For CaF2;

Number of moles reacted= mass/molar mass

Molar mass of CaF2= 78.07 g/mol

Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride

Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride

0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride

Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass

Molar mass of hydrogen fluoride= 20.01 g/mol

Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)

Actual yield of HF was given in the question as 2.2g

% yield of HF= actual yield/ theoretical yield ×100

%yield of HF= 2.2/5.6 ×100

% yield of HF= 39.3%

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6Ce^{4+} + I^- + 6OH^- → 6Ce^{3+} + IO_3^- + 3H_2O is the balanced chemical equation.

<h3>What is a balanced chemical equation?</h3>

A balanced chemical reaction is an equation that has equal numbers of each type of atom on both sides of the arrow.

Half-reaction method:

Unbalanced chemical equation:

Ce^{4+} + I^-→ Ce^{3+} + IO^{3-}

Oxidation half-reaction:

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Reduction half-reaction:

Ce4^+ + e^- → Ce^{3+}

Balanced chemical equation:

6Ce^{4+} + I^- + 6OH^-→ 6Ce^{3+} + IO^{3-} + 3H_2O

Oxidation number method:

Unbalanced chemical equation:

Ce^{4+} + I^-→ Ce^{3+} + IO^{3-}

I^{-1} -6e^-→ I^{+5}

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Balanced chemical equation:

6Ce^{4+} + I^{-1} → 6Ce^{3+} + I^{+5}

or

6Ce^{4+} + I^- + 6OH^-→ 6Ce^{3+} + IO_3^- + 3H_2O

Hence, 6Ce^{4+} + I^- + 6OH^-→ 6Ce^{3+} + IO_3^- + 3H_2O is the balanced chemical equation.

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https://brainly.in/question/46754758

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Answer:

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Explanation:

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