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Dmitry [639]
3 years ago
13

1. Determine the volume of SO2 (at STP) formed from the reaction of 96.7 mol FeS2 and 55.0 L of O2 at 358 K and 1.20 atm.

Chemistry
1 answer:
vlabodo [156]3 years ago
8 0

Answer:

40.0L of SO2 are produced

Explanation:

To solve this question we need to find the moles of O2 using PV = nRT in order to find the moles. Thus, we can find the limiting reactant and the moles (And volume) of SO2 produced as follows:

<em>Moles O2:</em>

n = PV/RT

n = 1.20atm*55.0L / 0.082atmL/molK*358K

n = 2.25 moles of O2.

Clearly, limiting reactant is O2.

The moles of SO2 produced are:

2.25 moles of O2 * (8mol SO2 / 11mol O2) = 1.6351 moles SO2

<em>Volume SO2:</em>

V = nRT/P

V = 1.6351 moles SO2*0.082atmL/molK*358K / 1.20atm

V = 40.0L of SO2 are produced

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Based on this, if there are two atoms, each with 6 protons but one of these is negatively charged it is because the second atom has 1 or more extra.

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two kilogram of nasialo4 zeolites are used in a water filter to remove calcium. after complete ion exchange the zeolite is remov
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After ion exchange with zeolite in water filter to remove calcium by substitute natrium ion with calcium ion, the weight of zeolite become 1.9578 kg.

ion exchange calculation can be done by using basic stoichiometry if we know how the reaction of the ion exchange. In this case where zeolite with natrium ion will be exchanged with calcium. The reaction of exchange will be like this:

2NaSiAlO_{4} + Ca^{2+} ⇒ Ca(SiAlO_{4}) _{2} + 2Na^{+}

Calculate the molecule weight of both zeolite
Atomic weight data:
Na = 23 gram/mole

Si = 28 gram/mole

Al = 27 gram/mole

O = 16 gram/mole

Ca = 40 gram/mole

Molecule weight

NaSiAlO_{4} = 142 gram/mole

Ca(SiAlO_{4}) _{2} = 278 gram/mole

Calculate mole from 2kg of zeolite with natrium :

2000 gram/ 142 = 14.0845 mole

Based on the stoichiometry, we got the mole of zeolite with calcium:

14.0845 mole x 1/2 = 7.0423 mole

Weight of zeolite after the ion exchanged with calcium:

7.0423 mole x 278 = 1957.8 gram = 1.9578 kg

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