All categories

3 years ago

Read the scenario.

Chemistry

1 answer:

bezimeni [28]3 years ago

4 0

The chemical equation of the synthesis reaction of ammonia is

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

Explanation:

Molar ratio is the ratio of number of moles between any two elements in the reactants side or in between reactant and product side.

Like, in the present situation, it is stated that the molar ratio between hydrogen gas and nitrogen gas is 3 : 1. This means 3 number of hydrogen molecules and 1 number of nitrogen molecule are required to form the product ammonia.

Since, it is known that two hydrogen atoms will make a single hydrogen molecule, so 3 number of hydrogen molecules means, 3 H₂ is one of the reactant. Similarly, two nitrogen atoms can make a single nitrogen molecule, so N₂ is another reactant. Thus the two reactants are obtained as 3 H₂ and N₂.

Now, it is also stated that the molar ratio between ammonia (NH₃) and hydrogen gas (H₂) is 2 : 3. This means 2 number of ammonia molecules are present in the product side. And we know that 3 number of hydrogen molecules are present in the reactant side.

Thus, the chemical equation of the synthesis reaction of ammonia is

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

You might be interested in

Help! giving brainly if correct

Gnesinka [82]

Answer:

I would say the answer is b.

Explanation:

6 0

2 years ago

Read 2 more answers

Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

$\delta H^0__{R,T_2}$ = $\delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'$

where:

$\delta H^0__{R}$ = enthalpy of reaction

${\delta C_p(T')}$ = the difference in the heat capacities of the products and the reactants.

∴

$\delta H^0__{R,435K}$ = $\delta H^0__{R,298.15K}$ $+$ $\int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'$

= $1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'$

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

$\delta H^0__{R,435K}$ ≅ 91383 J

6 0

3 years ago

PLZ HELP!!!!

Greeley [361]

In oil and gas industry:
When crude oil get extracted from well, salt water and some other stuff needs to be removed before oil can be sued in the car

6 0

3 years ago

How many molecules are there in 0.45 moles of so3

vivado [14]

Answer:

2.71

Explanation:

6 0

2 years ago

If a chemical reaction goes to completion, what can we say about the equilibrium constant for that reaction?

aniked [119]

Answer:

A. The equilibrium constant is very large

Explanation:

The equilibrium constant value is the ratio of the concentrations of the products over the reactants. When a chemical reaction goes to completion, that means that all the reactant has turned into products. As the equilibrium constant defines, it is the ratio of the product to the reactant. So at the final stage of the chemical reaction, the equilibrium constant will be very large.

7 0

3 years ago