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3 years ago

Level 1: 2Sr + 02 > 2Sro

Chemistry

1 answer:

tekilochka [14]3 years ago

8 0

Answer:

Sr would be the limiting reactant

5 moles

Explanation:

Since the equation is a balanced equation, the coefficient shows how each substance relates to the other in terms of the number of moles.

Reactants would be those on the left hand side of the arrow, while the products would be found on te right and side of the arrow. In this question, the reactants would be Sr and O₂.

Limiting reactant is the reactant that is insufficient; meaning to say that there is not enough of that substance and thus the reaction cannot continue. The other reactant(s) that is not limiting is called the excess reactants.

From the balanced equation, 2 moles of Sr is needed to react with 1 mole of O₂. Thus, if we have 5 moles of each reactant, Sr would be the limiting reactant since for every 1 mole of O₂, there has to be 2 moles of Sr in order for the reaction to proceed. Thus, if we have 5 moles of O₂, we would need 10 moles of Sr.

When we work out the amount of products formed, we look at the number of moles of the limiting reactant. This is because the limiting reactant determines how much is being reacted, while the excess number of moles of the excess reactant will remain unreacted.

For every 2 moles of Sr reacted, 2 moles of SrO would be produced. This means that the mole ratio of Sr to SrO is 1:1. Thus, since 5 moles of Sr has been reacted, 5 moles of the product (SrO) would be produced.

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At 311 K, this reaction has a K c value of 0.0111 . X ( g ) + 2 Y ( g ) − ⇀ ↽ − 2 Z ( g ) Calculate K p at 311 K. Note that the

Aleks [24]

Answer:

$K_{p}=4.35\times 10^{-4}$

Explanation:

We know, $K_{p}=K_{c}(RT)^{\Delta n}$

where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and $\Delta n$ is difference in sum of stoichiometric coefficient of products and reactants

Here $\Delta n=(2)-(2+1)=-1$ and T = 311 K

So, $K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}$

Hence value of equilibrium constant in terms of partial pressure $(K_{p})$ is $4.35\times 10^{-4}$

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3 years ago

Which of the following best describes the particles in an atom?

yanalaym [24]

Answer:

Electrons move around a nucleus.

Explanation:

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2 years ago

If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma

Veseljchak [2.6K]

Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

= 29.07 gram of Cu(NO3)2

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3 years ago

The following charges on individual oil droplets were obtained during an experiment similar to Millikan's. Determine a charge fo

RSB [31]

Answer:

- 1.602 x 10⁻¹⁹coulombs

Explanation:

Charge on individual oil droplet would be multiple of charge on one electron . So we will find out the minimum common factor of given individual charges that is the LCM of all the charges given.

LCM of given charges like 3.204 , 4.806 ,8.01 and 14.42 . We have neglected the power of ten( 10⁻¹⁹) because it is already a common factor to all.

The LCM is 1.602 . So charge on electron is 1.602 x 10⁻¹⁹.

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3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$