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1 year ago

If 5 grams of co and 5 grams of o2 are combined according to the reaction 2co o2 --> 2co2, which is the limiting reagent?

1 answer:

8 0

When 5 grams of CO and 5grams of O₂ is combined according to the reaction 2CO + O₂ ----------> 2CO₂ then Limiting reagent is CO.

Limiting reagent is the reactant that get used up in the reaction first.

According to the given reaction:

2CO + O₂ ----------> 2CO₂

5g 5g

∴ Molar mass of CO = Molar mass of C + Molar mass of O

⇒ Molar mass of CO = 14 + 16

⇒ Molar mass of CO = 28g

∴ Molar mass of O₂ = 16(2) = 32g

∴ Molar mass of CO₂ = Molar mass of C + 2(Molar mass of O)

⇒ Molar mass of CO₂ = 14 + 2(16)

⇒Molar mass of CO₂ =44g

Let's find out the moles of CO and O₂

∴ Moles = Given mass / Molar mass

⇒ moles of CO = 5/28 = 0.17

⇒ moles of O₂ = 5/32 = 0.15

For finding out the Limiting Reagent, we will divide the number of moles with the stiochiometry of the given reaction.

⇒ For CO = Moles/ stiochiometry = 0.17/2 = 0.085

⇒ For O₂ = Moles/ stiochiometry = 0.15/1 = 0.15

Since, the ratio of number of moles with the stiochiometry is less for CO hence it is the Limiting reagent, i.e. it will get used up in the reaction first.

Hence, the Limiting reagent for the reaction is CO.

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Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): 2H2O2

zalisa [80]

Answer:

23.0733 L

Explanation:

The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

$Mass=\frac {50}{100}\times 125\ g$

Mass = 62.5 g

Molar mass of $H_2O_2$ = 34 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{62.5\ g}{34\ g/mol}$

$moles= 1.8382\ mol$

Consider the given reaction as:

$2H_2O_2_{(aq)}\rightarrow2H_2O_{(l)}+O_2_{(g)}$

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.

Also,

1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.

So,

1.8382 moles of hydrogen peroxide decomposes to give $\frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr
The conversion of P(torr) to P(atm) is shown below:
[tex]P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 746 / 760 atm = 0.9816 atm

Temperature = 27 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (27 + 273.15) K = 300.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K

⇒V = 23.0733 L

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If 12 moles of Al2O3 are produced, how many moles of aluminum reacted?

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Explanation:

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