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Ivan
4 years ago
7

How many grams of water can be heated 20C by the addition of 1000J?

Chemistry
1 answer:
nydimaria [60]4 years ago
4 0
I hope this is correct

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Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a
Luden [163]

Answer : The mass of NaN_3 at temperature 20^oC 28.47 g.

The mass of NaN_3 at temperature 10^oC 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = 20^oC=273+20=293K     (0^oC=273K)

Volume of gas = 25\times 25\times 20cm=12500cm^3=12.5L   (1L=1000cm^3)

Molar mass of NaN_3 = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature 20^oC. The gas produced in the given reaction is N_2.

Using ideal gas equation,

PV=nRT

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)

By rearranging the terms, we get the value of 'n'

n=0.7015moles

The moles of N_2 = 0.7015 moles

The given balanced reaction is,

20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)

As, 32 moles of N_2 produced from 20 moles of NaN_3

So, 0.7015 moles of N_2 produced from \frac{20}{32}\times 0.7015=0.438 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

\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g

Therefore, the mass of NaN_3 needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature 10^oC and same volume & pressure.

Using ideal gas equation,

PV=nRT

Now put all the given values in this formula, we get

(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)

By rearranging the terms, we get the value of 'n'

n=0.726moles

The moles of N_2 = 0.726 moles

The given balanced reaction is,

20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)

As, 32 moles of N_2 produced from 20 moles of NaN_3

So, 0.726 moles of N_2 produced from \frac{20}{32}\times 0.726=0.454 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

\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g

Therefore, the mass of NaN_3 needed are 29.51 g.

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