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3 years ago

How many grams of h2 are needed to produce 10.78 g of nh3?

1 answer:

6 0

Ammonia is formed by a reaction between hydrogen and nitrogen as shown by the equation below.
N2(g) + 3H2(g) = 2NH3(g)
1 mole of ammonia contains 17 g
Therefore 10.78 g of ammonia are equivalent to 10.78/17 = 0.6341 moles
The mole ratio of hydrogen to ammonia is 3 : 2
Therefore, moles of hydrogen used will be 0.6341 × 3/2 = 0.9512 moles
1 mole of hydrogen is equivalent to 2 g
Thus, the mas of hydrogen will be 0.9512 moles × 2 = 1.9023 g

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Aqua regia, a mixture of HCl and HNO₃, has been used since alchemical times to dissolve many metals, including gold. Its orange

marissa [1.9K]

Aqua regia is an oxidative mixture that is highly corrosive and is composed of hydrochloric acid and nitric acid. The Ea (rev) for the reaction is 3 kJ.

<h3>What is activation energy?</h3>

The activation energy is the minimum required energy by the reactant to undergo changes to form the product. The activation energy of the reverse reaction is given by the difference in the production state and transition state.

It is given as,

Ea (rev) = Ea (fwd) − ΔHrxn

Given,

ΔH° = 83KJ

Ea (fwd) = 86 kJ/mol

Substituting the values above as:

Ea (rev) = 86 - 83

= 3 kJ

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4 0

1 year ago

Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

3 years ago

A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam

Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0

2 years ago