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If 107g of K2SO4 is used in the reaction 2 AlBr3 + 3 K2SO4 → 6 KBr + 1 Al2(SO4)3, how many particles of Al2(SO4)3 will be produced? Express your answer in scientific notation, for example, 6.02e23

User IVNSTN
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Final answer:

The number of particles of Al2(SO4)3 produced can be calculated by converting the given mass of K2SO4 to moles, and then using the mole ratio to determine the moles of Al2(SO4)3. Finally, the number of particles can be determined using Avogadro's number. Number of particles of Al2(SO4)3 = Moles of Al2(SO4)3 * Avogadro's number = 0.2049 mol * 6.022 × 10^23 particles/mol = 1.234 × 10^23 particles.

Step-by-step explanation:

In the given reaction, 2 AlBr3 + 3 K2SO4 → 6 KBr + 1 Al2(SO4)3, the molar ratio between K2SO4 and Al2(SO4)3 is 3:1. Therefore, for every 3 moles of K2SO4 used, 1 mole of Al2(SO4)3 is produced.

First, convert the given mass of K2SO4 to moles using its molar mass. The molar mass of K2SO4 is 174.26 g/mol.

Moles of K2SO4 = Mass of K2SO4 / Molar mass of K2SO4 = 107 g / 174.26 g/mol = 0.6148 mol

Since the mole ratio is 3:1, the number of moles of Al2(SO4)3 produced is 1/3 of the moles of K2SO4 used.

Moles of Al2(SO4)3 = Moles of K2SO4 * (1/3) = 0.6148 mol * (1/3) = 0.2049 mol

To calculate the number of particles, we can use Avogadro's number. There are 6.022 × 10^23 particles in 1 mole of any substance.

Number of particles of Al2(SO4)3 = Moles of Al2(SO4)3 * Avogadro's number = 0.2049 mol * 6.022 × 10^23 particles/mol = 1.234 × 10^23 particles

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