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What is the maximum mass of PH3 that can be formed when 62.0g of phosphorus reacts with

Luda [366]

Answer: The mass of $PH_3$ produced is 45.22 g

Explanation:

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.

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

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

For $P_4$ :

Given mass of $P_4$ = 62.0 g

Molar mass of $P_4$ = 124 g/mol

Putting values in equation 1, we get:

$\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol$

For $H_2$ :

Given mass of $H_2$ = 4.00 g

Molar mass of $H_2$ = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol$

The chemical equation follows:

$P_4(g)+6H_2(g)\rightarrow 4PH_3(g)$

By stoichiometry of the reaction:

If 6 moles of hydrogen gas reacts with 1 mole of $P_4$

So, 2 moles of hydrogen gas will react with = $\frac{1}{6}\times 2=0.333mol$ of $P_4$

As the given amount of $P_4$ is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 6 moles of $H_2$ produces 4 mole of $PH_3$

So, 2 moles of $H_2$ will produce = $\frac{4}{6}\times 2=1.33mol$ of $PH_3$

We know, molar mass of $PH_3$ = 34 g/mol

Putting values in equation 1, we get:

$\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g$

Hence, the mass of $PH_3$ produced is 45.22 g

6 0

3 years ago

Sodium bromide+ Lithium Based on Single replacement

Helga [31]

Answer:

the answer will be lithium bromide + sodium

Explanation:

because displace sodium from bromine there by liberating sodium as an element in the product side

6 0

3 years ago

What do chemist study

enyata [817]

Answer:

Chemists study matter and the chemical reactions between substances

Explanation:

8 0

3 years ago

Read 2 more answers

What is the total amount of heat required to completely melt 347 grams of ice at its melting point?

Bogdan [553]

In order to calculate the amount of heat energy required to melt 347 grams of ice, we need to will apply the equation:
Q = ml, where m is the mass of substance and l is its latent heat of fusion. For ice, the latent heat of fusion is 334 joules per gram. Therefore:
Q = 347 x 334
Q = 112,558 joules of energy

Hope this helped.

7 0

3 years ago

A 5.00g piece of metal is heated to 100.0°C, then placed in a beaker containing 20.0 of water at 10.0°C. The temperature of the

Ahat [919]

Answer:

This metal has a specific heat of 0.9845J/ g °C

Explanation:
Step 1: Given data
q = m*ΔT *Cp
⇒with m = mass of the substance
⇒with ΔT = change in temp = final temperature T2 - initial temperature T1
⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)

Step 2: Calculate specific heat
For this situation : we get for q = m*ΔT *Cp
q(lost, metal) = q(gained, water)

- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)
-5 * (15-100)(Cpmetal) = 20* (15-10) * (4.184J/g °C =
-5 * (-85)(Cpmetal) = 418.4

Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C

This metal has a specific heat of 0.9845J/ g °C

7 0

3 years ago