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3 years ago

Identify the catalyst in this reaction, explain how

Chemistry

1 answer:

telo118 [61]3 years ago

5 0

Question:

Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:

equations

Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.

Answer:

It is present but not consumed

NO Lowers the activation energy of the reaction

Explanation:

A catalyst is a substance that is present in a chemical reaction and enables the reaction to occur at a faster rte but does not take part n the reaction

Therefore, whereby NO is not consumed, it is the catalyst

It functions by lowering the activation energy

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2 years ago

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The difference between the density of gas and the density of liquid

eduard

The molecules of a liquid substance are closely packed together to each other. So as a result, liquids are denser than gases.

<h3>What is the difference between the density of liquid and gas?</h3>

A mass of gas will have a much larger volume compared to the same mass of liquid. This is because it has a much lower density. The density of gaseous oxygen is 0.0014 g/cm3. Density is ρ=Mass Volume. We know that gas will uniformly occupy more space than liquid whatever volume is available to it. On the other hand, solids and liquids, are closely packed as compared to gas and are high-density materials where ρ is relatively constant.

So we can conclude that the molecules of a liquid substance are closely packed together with each other. So as a result, liquids are denser than gases.

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1 year ago

Heat is which of the following???

Sergeu [11.5K]

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3 years ago

I need help solving this!

zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol$

The given reaction equation is as follows.

$C + 2H_{2} \rightarrow CH_{4}$

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

$Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol$

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

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3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago