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Wittaler [7]
3 years ago
5

. Which substance is the limiting reactant when 2 g of S reacts with 3 g of O2 and 4 g of NaOH according to the following chemic

al equation: 2 S (s) + 3 O2 (g) + 4 NaOH (aq) → 2 Na2SO4 (aq) + 2 H2O (l) a) S (s) b) O2 (g) c) NaOH (aq) d) none of these substances is the limiting reactant.
Chemistry
1 answer:
Ksivusya [100]3 years ago
3 0

Answer:

S is the limiting reagent.

Explanation:

To find the limiting reactant we must first write the balanced chemical reaction. It must be correctly balanced so that we can find the proper mole ratios.

2 S (s) + 3 O2 (g) + 4 NaOH (aq) → 2 Na2SO4 (aq) + 2 H2O (l)

After this we will convert our measurements to moles. For mass we do this by dividing by the molar mass.

2g ÷ 32.06 = 0.06238mol S

3g ÷ 32.00 = 0.09375mol O₂

Now that we have the moles of each of the reactants, we can multiply them by their mole ratio with a reactant.

0.06238mol S × 2/2 = 0.06238mol H2O

0.09375mol O₂ × 2/3 = 0.06250mol H2O

S is our limiting reagent because it makes the smaller amount of moles.

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Answer:

28.11g

Explanation:

Step 1:

Background understanding:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10^23 atoms. This also indicates that 1 mole of helium (He) contains 6.02x10^23 atoms.

1 mole of He = 4g

Step 2:

Determination of the mass of He that contain 4.23x10^24 atoms. This is illustrated below:

4g of He contains 6.02x10^23 atoms.

Therefore Xg of He will contain 4.23x10^24 atoms i.e

Xg of He = (4x4.23x10^24)/6.02x10^23

Xg of He = 28.11g

Therefore, 28.11g of He contains 4.23x10^24 atoms

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