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Tcecarenko [31]
3 years ago
12

The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces

large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide ( NaN3 ) into solid sodium and gaseous dinitrogen. 2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm . Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.
Chemistry
1 answer:
Oxana [17]3 years ago
5 0

Answer:

1. NaN₃(s) → Na(s) + 1.5 N₂(g)

2. 79.3g

Explanation:

<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>

NaN₃(s) → Na(s) + 1.5 N₂(g)

<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>

First, we have to calculate the moles of N₂ from the ideal gas equation.

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1atm.(43.0L)}{(0.08206atm.L/mol.K).286.2K} =1.83mol

The moles of NaN₃ are:

1.83molN_{2}.\frac{1molNaN_{3}}{1.5molN_{2}} =1.22molNaN_{3}

The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:

1.22mol.\frac{65.01g}{mol} =79.3g

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3 0
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IrinaVladis [17]

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3 0
3 years ago
The temperature inside your freezer is 0 degrees Celsius. You place a balloon with an initial temperature of 30 degrees C and a
blsea [12.9K]

Answer:

V=0.68L

Explanation:

For this question we can use

V1/T1 = V2/T2

where

V1 (initial volume )= 0.75 L

T1 (initial temperature in Kelvin)= 303.15

V2( final volume)= ?

T2 (final temperature in Kelvin)= 273.15

Now we must rearrange the equation to make V2 the subject

V2= (V1/T1) ×T2

V2=(0.75/303.15) ×273.15

V2=0.67577931717

V2= 0.68L

4 0
3 years ago
What is the <br> formula for gunpowder
mrs_skeptik [129]

Here is the chemical equation for gun powder, in it’s simple form:

<span>2 KNO3 + S + 3 C → K2S + N2 + 3 CO2.</span>

This is the same simplified formula, only balanced:

<span>10 KNO3 + 3 S + 8 C → 2 K2CO3 + 3 K2SO4 + 6 CO2 + 5 N2.</span>

3 0
3 years ago
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
3 years ago
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