10 KClO3 (aq) + 3 P4 (s) --> 3 P4O10 (s) + 10 KCl (s) , what mass of P4O10 must be produced if 2.00x10^23 particles of KClO3 are also consumed

Answer 1

10 KClO₃ (aq) + 3 P₄ (s) --> 3 P₄O₁₀ (s) + 10 KCl (s) ... balanced equation

2.00x10²³ particles KClO₃ x 1 mol KClO₃ / 6.02x10²³ particles = 0.332 mols KClO₃ consumed

0.332 mols KClO₃ x 3 mol P₄O₁₀ / 10 mol KClO₃ x 284 g P₄O₁₀ / mol = 28.3 g P₄O₁₀

Answer 2

You want to convert your known number to moles first. This can be done by converting from particles to moles. We can use Avogadro's constant, 6.022*10^23, as there is that number of particles in one mole of any atom or molecule.

(2.00*10^23 particles)/(6.022*10^23 particles per mole) = 0.3321155762 moles of KClO3

From the equation, for every 10 moles of KClO3 consumed, there are 3 moles of P4O10 produced.

So,

0.3321155762 mol KClO3 * (3 mol P4O10/10 mol KClO3) = 0.09963467286 mol P4O10

Then, converting from moles to grams, the molar mass of P4O10 is 4(30.97) + 16(10) = 283.88 g/mol

0.09963467286 mol * 283.88 g/mol = 28.2842909315

To stay consistent with sig figs, we need to reduce this number to 3 sig figs.

The final answer is: 28.3 grams of P4O10 are produced.