Write the complete balanced chemical equation _A+_BC3->_AC2 +

Now we need to find the amount of NF3 that can be formed by the complete reactions of each of the reactants. If all of the N2 wa

Alex777 [14]

The question is incomplete, the complete question is:

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

$N_2(g)+3F_2(g)\rightarrow 2NF_3(g)$

A sample contains 19.3 g of $N_2$ is reacted with 19.3 g of $F_2$ . Now we need to find the amount of $NF_3$ that can be formed by the complete reactions of each of the reactants.

If all of the $N_2$ was used up in the reaction, how many moles of $NF_3$ would be produced?

Answer: 1.378 moles of $NF_3$ are produced in the reaction.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

In the given chemical reaction, $N_2$ is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.

We are given:

Mass of $N_2$ = 19.3 g

Molar mass of $N_2$ = 28.02 g/mol

Putting values in equation 1:

$\text{Moles of }N_2=\frac{19.3g}{28.02g/mol}=0.689mol$

For the given chemical reaction:

$N_2(g)+3F_2(g)\rightarrow 2NF_3(g)$

By the stoichiometry of the reaction:

1 mole of $N_2$ produces 2 moles of $NF_3$

So, 0.689 moles of $N_2$ will produce = $\frac{2}{1}\times 0.689=1.378mol$ of $NF_3$

Hence, 1.378 moles of $NF_3$ are produced in the reaction.

8 0

3 years ago

An atom that has lost one or more electrons is a(n) _______

kompoz [17]

Answer:

An atom that loses one or more valence electrons to become a positively charged ion is known as a cation, while an atom that gains electrons and becomes negatively charged is known as an anion.

8 0

3 years ago

Read 2 more answers

1.Which of the following are considered pure substances

PilotLPTM [1.2K]

1.) homogeneous mixtures

2.) NaCI, because sodium is a metal and chlorine is a nonmetal

3.) I'm thinking the first the option

5 0

3 years ago

Chlorine and bromine are elements in Group VII.

Goryan [66]

Explanation:

Bromine is a chemical element with the symbol Br and atomic number 35. It is the third-lightest halogen, and is a fuming red-brown liquid at room temperature that evaporates readily to form a similarly coloured vapour. Its properties are intermediate between those of chlorine and iodine. Isolated independently by two chemists, Carl Jacob Löwig (in 1825) and Antoine Jérôme Balard (in 1826), its name was derived from the Ancient Greek βρῶμος ("stench"), referring to its sharp and disagreeable smell.

Bromine, 35Br

7 0

3 years ago

Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.

Nastasia [14]

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

5 0

3 years ago

Write the complete balanced chemical equation _A+_BC3->_AC2 +_ B