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Alex777 [14]
2 years ago
10

According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams

of chlorine gas? 3 C12(g) + 3 H2O(1) ► 5 HCl(aq) + HCO3(aq) grams HC103
Chemistry
1 answer:
gogolik [260]2 years ago
4 0

Answer:

m_{HClO_3}=12.7gHClO_3

Explanation:

Hello,

Considering the reaction:

3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3

The molar masses of chlorine and chloric acid are:

M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol

Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:

m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3

Best regards.

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How many molecules of CBr4 are in 250 grams of CBr4
Kazeer [188]

Answer:- 4.54*10^2^3 molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


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2 years ago
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Answer:

Explanation:

final temperature of the cube

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4 0
2 years ago
Oxides of nitrogen are pollutant gases which are emitted from car exhausts.
V125BC [204]
Let's look at the molar weight of the answers: 
NO is 30 g/mol 
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<span>We have the grams of the product, so we need the moles in order to calculate the molar weight. We us PV=nRT for this, assuming standard temperature and pressure. </span>
You were given the liters (.120L) 
Std pressure is 1 atmosphere 
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<span>Temp is 293.15 kelvin, thats standard </span>
And r is the gas constant in liters-atm per mol kelvin 

(.120 liters)(1atm)=n(293.15K)(.08206) 
Solving for n is .0049883835 mol 

<span>.23g divided by .0049883 mol is about 46g&#x2F;mol. You&#x27;re answer is B I think, NO2

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4 0
2 years ago
Read 2 more answers
What element is being oxidized in the following redox reaction?
gregori [183]

Answer:

C is the element thats has been oxidized.

Explanation:

MnO₄⁻ (aq)  +  H₂C₂O₄ (aq)  →  Mn²⁺ (aq)  +  CO₂(g)

This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.

In the oxalic acid, this are the oxidation states:

H: +1

C: +3

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In the product side, in CO₂ the oxidation states are:

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4 0
2 years ago
The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3
Oksanka [162]

Answer : The mass of NaN_3 required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = 113 L

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 85^oC=273+85=358K

Putting values in above equation, we get:

1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K

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Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 3 mole of N_2 produced from 2 mole of NaN_3

So, 3.84 moles of N_2 produced from \frac{2}{3}\times 3.84=2.56 moles of NaN_3

Now we have to calculate the mass of NaN_3

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

Molar mass of NaN_3 = 65 g/mole

\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g

Therefore, the mass of NaN_3 required is, 166.4 grams.

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