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3 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]3 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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an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

3 years ago

Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have

Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0

3 years ago

Katrina tried to take pictures with her cell phone but discovered that its battery was dead. What happened to the battery?

Aleks04 [339]

C. All the energy stored in the battery had been transformed.

4 0

2 years ago

Read 2 more answers

Not enough air is pumped into an inflatable life raft to make completely filled out the atm pressure is 14.7psi what is the pres

sasho [114]

If you do not inflate the life raft to make completely filled out, as long as you do not press or squeeze the life raft, the air inside it will be in equilibrium with the air outside the raft, and so the pressure inside the life raft will be the same atmospheric pressure, 14.7 psi.

Note that when the raft is swollen, if you punch it, the air will leave from it which means that the pressure inside is greater than the atmospheric pressure.

5 0

3 years ago

If a recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb what volume in liters

Bad White [126]

The volume in liters of 3245 aluminum cans is 24.8 L.

A recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb. The mass of 3245 aluminum cans is:

$3245 cans \times \frac{1lb}{22cans} = 147.5 lb$