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3 years ago

35.0 grams of nitrogen gas reacts with 60.0 grams of hydrogen gas: N2 + 3H2--> 2NH3

Chemistry

1 answer:

Misha Larkins [42]3 years ago

6 0

Explanation:

Moles of N2 = 35.0g / (28g/mol) = 1.25mol

Moles of H2 = 60.0g / (2g/mol) = 30.0mol

Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.

Moles of NH3 = 1.25mol * 2 = 2.50mol.

Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.

30.0mol - 1.25mol * 3 = 26.25mol.

Excess mass of H2

= 26.25mol * (2g/mol) = 52.5g.

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29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.

arsen [322]

The answer for the following mention bellow.

Therefore the final temperature of the gas is 260 k

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 150.0 kPa

Final pressure ( $P_{2}$ ) = 210.0 kPa

Initial volume ( $V_{1}$ ) = 1.75 L

Final volume ( $V_{2}$ ) = 1.30 L

Initial temperature ( $T_{1}$ ) = -23°C = 250 k

To find:

Final temperature ( $T_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

We know;

$\frac{P*V}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ × $\frac{V_{1} }{V_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $T_{1}$ ) represents the initial temperature of the gas

( $T_{2}$ ) represents the final temperature of the gas

So;

$\frac{150 * 1.75}{210 * 1.30}$ = $\frac{260}{T_{2} }$

( $T_{2}$ ) =260 k

Therefore the final temperature of the gas is 260 k

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2 years ago

At what temperature (in C) will a sample of gas occupy 91.3 L if it occupies 45.0 L at 70.0°C? Assume constant pressure.)

sdas [7]

Solution is here,

for initial case,

temperature(T1)=70°C=70+ 273=343K

vloume( V1) =45 L

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temperature( T2)=?

volume(V2)= 91.3 L

at constant pressure,

V1/V2 = T1/T2

or, 45/91.3 = 343/ T2

or, T2= (343×91.3)/45

or, T2=695.9 K = (695.9-273)°C=422.9°C

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