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2 years ago

How many moles of KCIO3 are in 5 g of KClO3?

2 answers:

7 0

1. Calculate the molar mass of KClO3: 39.098+ 35.453 + 3(16) = 122.551 g KClO3
2. 5gKClO3 x 1mol KClO3/ 122.551g KClO3 = 0.041mol of KClO3

STatiana [176]2 years ago

6 0

Molar mass of KClO3

$\\ \tt\hookrightarrow 39+35+3(16)$

$\\ \tt\hookrightarrow 74+48$

$\\ \tt\hookrightarrow 122g/mol$

Now

$\\ \tt\hookrightarrow No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \tt\hookrightarrow No\:of\:moles=\dfrac{5}{122}$

$\\ \tt\hookrightarrow No\:of\:moles=0.04mol$

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2 years ago

What is the mass of a sample of alcohol (specific heat = 2.4 J/gC), if it requires 4780 J of heat to raise the temperature by 5.

insens350 [35]

The mass of a sample of alcohol is found to be = m = 367 g

Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.

Explanation:

The Energy of the sample given is q = 4780

We are required to find the mass of alcohol m = ?

Given that,

The specific heat given is represented by = c = 2.4 J/gC

The temperature given is ΔT = 5.43° C

The mass of sample of alcohol can be found as follows,

The formula is c = $\frac{q}{mt}$

We can drive value of m bu shifting m on the left hand side,

m = $\frac{q}{ct}$

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3 years ago

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

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