All categories

3 years ago

Use the reaction below for the decomposition of sodium azide

Chemistry

1 answer:

bezimeni [28]3 years ago

8 0

<u>Answer:</u> The volume of nitrogen gas at STP is 44.8 L.

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation for it follows:

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

We are given:

Given mass of $NaN_3$ = 130.0 g

Molar mass of $NaN_3$ = 65.01 g/mol

Using equation 1:

$Moles of NaN_3=\frac{130.0 g}{65.01g/mol}=2mol$

For the given chemical equation:

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

By the stoichiometry of the reaction:

2 moles of $NaN_3$ produces 3 moles of nitrogen gas

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 2 moles of nitrogen gas will occupy $=\frac{22.4L}{1mol}\times 2mol=44.8L$ of volume

Hence, the volume of nitrogen gas at STP is 44.8 L.

You might be interested in

A large balloon contains 5400 m3 of He gas that is kept at a temperature of 280 K and an absolute pressure of 1.10 x 105 Pa. Fin

patriot [66]

Answer:

1.02 × 10⁶ g

Explanation:

Step 1: Given data

Volume of the balloon (V): 5400 m³

Temperature (T): 280 K

Absolute pressure (P): 1.10 × 10⁵ Pa

Molar mass of He (M): 4.002 g/mol

Step 2: Convert "V" to L

We will use the conversion factor 1 m³ = 1000 L.

5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 101325 Pa.

1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm

Step 4: Calculate the moles of He (n)

We will use the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K

n = 2.56 × 10⁵ mol

Step 5: Calculate the mass of He (m)

We will use the following expression.

m = n × M

m = 2.56 × 10⁵ mol × 4.002 g/mol

m = 1.02 × 10⁶ g

8 0

3 years ago

Son el punto 2,3,4 por favor

RSB [31]

4...........................

8 0

3 years ago

The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to P

expeople1 [14]

Answer : The ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

Solution : Given,

$K_p=1.5\times 10^6$

$p_{O_2}$ = 0.21 atm

The given equilibrium reaction is,

$NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}$

Now put all the given values in this expression, we get:

$1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}$

$\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}$

$\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5$

Therefore, the ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

5 0

3 years ago

Consider this reaction 2Mg(s)+O2(g) ———> 2MgO(s) What volume (in milliners) of gas is required to react with 4.03 g Mg at STP

nydimaria [60]

The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml

calculation/

calculate the moles of Mg used

moles=mass/molar mass

moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles

by use of mole ratio of Mg:O2 from the equation which is 2:1

the moles 02=0.1679 x1/20.0829 moles

at STP 1 mole of a gas= 22.4 l

0.0895 moles=? L

by cross multiplication

=0.0895 moles x22.4 l/ 1 mole=1.8570 L

into Ml = 1.8570 x1000=1856 ml approximately to 1860

6 0

3 years ago

PLS ANSWER ASAP

Kazeer [188]

The answer is c this is the answer

7 0

3 years ago

Read 2 more answers