What is the total number of moles contained in 115g of C2H5OH?

If 2.0 mol of Zn is mixed with 3.0 mol of HCl, which reactant will be limiting?

Soloha48 [4]

The following Balanced Reaction will take place:

Zn + 2HCl → ZnCl₂ + H₂

In the question, we have 2 moles of Zinc and 3 moles of HCl for this reaction

Amount of HCl required to completely react with 2 moles of Zn:

Since we need 2 moles of HCl for every mole of Zn, we will need 2(2) = 4 moles of HCl for every 2 moles of Zn

Identifying the Limiting Reagent:

But we are only given 3 moles of HCl where we need 4 moles to completely react.

So, since HCl is in less amount, it is the Limiting Reagent

8 0

2 years ago

Which is NOT a way nitrogen can be fixed or made more useful?

stich3 [128]

This doesn’t have a multiple choice part to answer.

7 0

2 years ago

Are nails in a jar with water a chemical change

Digiron [165]

I would say it is a chemical change because after a while, the nails start to rust. Now they only rust after about a week though. But rust is a chemical change.

7 0

2 years ago

1 2 3 4 5 6 7 8 9 10 The diagrams show gases that are stored in two separate but similar containers. 2 identical containers have

zvonat [6]

Answer:

gas 2 because it has more particles colliding.

7 0

3 years ago

Read 2 more answers

Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat

Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:

N: 14 g/mol
H: 1 g/mol
O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
O₂: 2*16 g/mol= 32 g/mol
NO: 14 g/mol + 16 g/mol= 30 g/mol
H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

NH₃: 4 moles
O₂: 5 moles
NO: 4 moles
H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

NH₃: 4 moles*17 g/mol= 68 g
O₂: 5 moles*32 g/mol= 160 g
NO: 4 moles*30 g/mol= 120 g
H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

$mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }$

mass of O₂≅17.65 g

17.65 grams of O2 are needed for a complete reaction.

3 0

2 years ago