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Alisiya [41]
3 years ago
5

How many grams of no2 are theoretically produced if we start with 1.20 moles of s and 9.90 moles of hno3?

Chemistry
1 answer:
Ierofanga [76]3 years ago
5 0

Answer:- Theoretically 7.20 moles of NO_2 are formed.

Solution:- The balanced equation for the reaction of Sulfur with nitric acid is---

S + 6HNO_3\rightarrow H_2SO_4 + 6NO_2 + 2H_2O

From balanced equation, 1 mol of sulfur reacts with 6 moles of nitric acid. So, 1.20 moles of sulfur would react with 6 x 1.20 = 7.20 moles of nitric acid.

9.90 moles of nitric acid are available. It means nitric acid is present in excess and sulfur is limited. So, the theoretical yield is calculated by 1.20 moles of sulfur as...

1.20 mol x (6 mol nitrogen dioxide/1mol S) = 7.20 mol NO_2

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This pertains to the concept that polar substances can dissolve only other polar substances. Also, nonpolar substances are also only able to dissolve nonpolar substances. 

Polarity of the substance depends primarily on the type of bond and the difference in electronegativity. 

Water is a polar substance while vegetable oil is not. From the concept presented above, it may be concluded that water will not be able to dissolve the vegetable oil and the assumption is logical. 
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"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
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<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

c_3=30.68J/g^oC

Hence, the specific heat of calorimeter is 30.68 J/g°C

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