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2 years ago

If 25.0 g NO are produced, how many grams of nitrogen gas are used?

1 answer:

8 0

Using the stoichiometry of the reaction and the information provided in the question, the mass of N2 used is 11.62 g.

<h3>Chemical reaction</h3>

The term chemical reaction refers to the combiantion of two or substances to yiled one or more products. The reaction equation in this case is N2 + O2 --->2NO.

Now;

Number of moles of NO = 25g/30 g/mol = 0.83 moles

1 mole of N2 yields 2 moles of NO

x moles of N2 yileds 0.83 moles of NO

x = 0.415 moles

Mass of N2 = 0.415 moles * 28 g/mol = 11.62 g

Learn more about stoichiometry:

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

<u>Answer:</u> The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

<u>Initial:</u> 0.025 0.025

<u>At eqllm:</u> 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

The paths in which electrons travel are called what ovals ,paths,circles,orbitals

PolarNik [594]

The paths in which electrons travel are called orbitals.

8 0

3 years ago

Read 2 more answers

Consider the equation ∆G = –nFE, where F = 9.648533289 x 104 C/mol. Given that n is the mole of e– transferred and the following

myrzilka [38]

Answer:

See explanation below.

Explanation:

In the equation ∆G = –nFE, E is the electromotive force ( cell potential ) in Volts.

Now in turn a Volt is defined as the potential difference that will impart one joule of energy per coulomb of charge that moves through two points.

V = J/C where J is Joules and C is coulombs of charge

Therefore in terms of units the equation will give us units of Joules:

[ mol] x [C/mol] x [J/C] = [J]

5 0

3 years ago

Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of a

kvv77 [185]

Answer:

The mass percent of Al(OH)₃ is 15.3%

Explanation:

The reaction is:

Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:

NaOH + HCl = NaCl + H₂O

The total moles of HCl is:

$n_{HCl,total} =M_{HCl} *V_{HCl} =0.111*0.025=2.78x10^{-3} moles$

From the second titration, the moles of excess of HCl is:

$n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles$

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

$n_{HCl,reacts} =n_{HCl,total}-n_{HCl,excess} =2.78x10^{-3} moles-1.46x10^{-3} moles=1.32x10^{-3} moles$

The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:

$m_{Al(OH)3} =1.32x10^{-3} molesHCl*\frac{1molAl(OH)3}{3molesHCl} *\frac{78gAl(OH)3}{1molAl(OH)3} =0.03g$

The percentage of Al(OH)₃ is:

$Percentage-Al(OH)3=\frac{m_{Al(OH)3} }{m_{antiacid} } *100=\frac{0.03}{0.196} =15.3$ %

3 0

3 years ago

CaCO3(s)⇄CaO(s)+CO2(g) When heated strongly, solid calcium carbonate decomposes to produce solid calcium oxide and carbon dioxid

rusak2 [61]

Answer:

1.04 mol

Explanation:

CO₂ is produced in a closed 100 L vessel according to the following equation.

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

At equilibrium, the pressure of carbon dioxide remains constant at 1.00 atm.

First, we need to conver the temperature to the absolute scale (Kelvin scale) using the following expression.

K = °C + 273.15

K = 898°C + 273.15

K = 1171 K

Now, we can find the moles of carbon dioxide using the ideal gas equation.

$P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1.00atm \times 100L}{\frac{0.0821atm.L}{mol.K} \times 1171K} = 1.04 mol$

7 0

3 years ago