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2 years ago

If 4500 g of Zr metal is reacted with 1.5 x 104 g of water what is the limiting reactant?

Chemistry

1 answer:

PIT_PIT [208]2 years ago

6 0

Answer:

Explanation:

The balance equation of the reaction is:

Zr + 2 H₂O ⇒ ZrO₂ + 2 H₂

Being:

Zr: 91 g/mole
H: 1 g/mole
O: 16 g/mole

the molar mass of the reactants participating in the reaction are:

Zr: 91 g/mole
H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

If by stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) they react in moles:

Zr: 1 mole
H₂O: 2 moles

then by stoichiometry the following quantity of mass react:

Zr: 1 mole*91 g/mole= 91 g
H₂O: 2 moles*18 g/mole= 36 g

Now the following rule of three applies: if 91 g of Zr react with 36 g of H₂O, 4500 g of Zr with how much mass of H₂O would it react?

$mass of H_{2} O=\frac{4500 grams of Zr*36 grams of H_{2}O }{91 grams of Zr}$

mass of H₂O= 1,780 grams

But 1780 grams of H₂0 are not available, 1.5 * 10⁴(15,000) grams are available. Since it has more mass than it needs to react with 4500 grams of Zr, Zr zirconia will be the limiting reagent.

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