Part A Which substance is the limiting reactant when 14 g of sulfur reacts with 24 g of oxygen and 28 g
of potassium hydroxide according to the following chemical equation?
2 S(s)+ 3 O2(g) + 4 KOH(q) — 2 K2SO4(aq) + 2 H20(1)
O O2(g)
S(s)
K2SO4
O KOH(aq)
O None of these substances is the limiting reactant.
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KOH is limiting reactant

Explanation:

To solve this question, we need to convert the mass of each reactant to moles. Then, using the balanced reaction, we can find the limiting reactant:

Moles S -Molar mass: 32g/mol-:

14g S * (1mol / 32g) = 0.44mol

Moles O₂ -Molar mass: 32g/mol-:

24g O₂ * (1mol / 32g) = 0.75mol

Moles KOH -Molar mass: 56g/mol-:

28g KOH * (1mol / 56g) = 0.5mol

For a complete reaction of the KOH are required:

0.5mol KOH * (2mol S / 4mol KOH) = 0.25mol S

0.5mol KOH * (3mol O₂ / 4mol KOH) = 0.38mol O₂

As there are 0.44 moles of S and 0.38moles of oxygen,

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the answer is d.indefinite shape but definite volume, let go oogle u out

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