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3 years ago

Calculate the theoretical yield of ammonia produced by the reaction of 100g of H2 gas and 200g of N2 gas

1 answer:

6 0

To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
= 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
mass of NH3 = no.moles * molar mass of ammonia
= 14.28 * 17 = 242.76 g

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If 60.2 grams of hg combines completely with 24.0 grams of br to form a compound what is the percent

NikAS [45]

Answer is: the percent composition of Hg in the compound is 71.5%.

Balanced chemical reaction: Hg + Br₂ → HgBr₂.

m(Hg) = 60.2 g; mass of the mercury.

m(Br₂) = 24.0; mass of the bromine.

m(HgBr₂) = m(Hg) + m(Br₂).

m(HgBr₂) = 60.2 g + 24 g.

m(HgBr₂) = 84.2 g; mass of the compound.

ω(Hg) = m(Hg) ÷ m(HgBr₂) · 100%.

ω(Hg) = 60.2 g ÷ 84.2 g · 100%.

ω(Hg) = 71.5%.

6 0

3 years ago

Read 2 more answers

What are the moles of gas are in a 30 liter scuba canister if the temperature of the canister is 300 K and the pressure is 200 a

Lady_Fox [76]

Answer:

$n = 243.605$

Explanation:

Given

$Pressure, P = 200\ atm$

$Temperature, T = 300k$

$Volume, V = 30L$

Required

Calculate the number of moles

We'll apply the following formula to solve this question

$n = \frac{pV}{RT}$

Where

$R = 0.0821 L\ atm/(mol\ K)$

The above equation is an illustration of the ideal gas law

Substitute values for p, V, R and T in:

$n = \frac{pV}{RT}$

$n = \frac{200 * 30}{0.0821 * 300}$

$n = \frac{6000}{24.63}$

$n = 243.605$

Hence, there are 243.605 moles

7 0

2 years ago

Explain how human activity can affect the carbon cycle.

Eva8 [605]

Answers:

Human activities badly affect carbon cycle. Activities such as burning fossil fuel and deforestation have begun to effect on carbon cycle and the rise of carbon dioxide in atmosphere.

Explanation:

The carbon cycle can be affect when carbon dioxide is either released into atmosphere or remove from atmosphere. When fossil are burnt , carbon is release to the atmosphere at faster rate then it is removed. Natural gas, oil, coal, and other industrial products all are affecting carbon cycle in atmosphere.

Deforestation means permanent removal of trees from forest which cause increase in level of carbon dioxide because no trees longer to absorb carbon dioxide from atmosphere. Which result affect on carbon cycle.

4 0

3 years ago

Read 2 more answers

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

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

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

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

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

3 years ago

What compound is formed when methyloxirane (1,2-epoxypropane) is reacted with ethylmagnesium bromide followed by treatment with

Pavlova-9 [17]

Answer:

Pentan-2-ol

Explanation:

On this reaction, we have a Grignard reagent (ethylmagnesium bromide), therefore we will have the production of a carbanion (step 1). Then this carbanion can attack the least substituted carbon in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the treatment with aqueous acid, when we add acid the hydronium ion ( $H^+$ ) would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be attacked by the negative charge produced in the second step to produce the final molecule: "Pentan-2-ol".

See figure 1

I hope it helps!

5 0

3 years ago