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3 years ago

2 CuCl2 +4KI -> 2 Cul + 4 KCI + 12

Chemistry

1 answer:

Amanda [17]3 years ago

5 0

Answer:

Explanation:

This is an example of a limiting reactant question, and is very common as a general chemistry problem.

We first see the balanced equation, that is:

2CuCl2+4KI→2CuI+4KCl+I2

We first need to find the limiting reactant

We see that 0.56 g of copper(II) chloride (CuCl2) reacts with 0.64 g of potassium iodide (KI) . So, let's convert those amounts into moles.

Copper(II) chloride has a molar mass of

134.45 g/mol . So in 0.56 g of copper(II) chloride, then there exist

0.56g134.45g/mol≈4.17⋅10−3 mol

Potassium iodide has a molar mass of

166 g/mol . So, in 0.64 g of potassium iodide, there exist

if it wrong i am sorry

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Read 2 more answers

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$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

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Putting values in above equation, we get:

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