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andreev551 [17]
3 years ago
14

A 43.63 gram sample of a hydrate of caseo4 was heated thoroughly in a porcelain crucible, until its weight remained constant. af

ter heating, 36.45 grams of the anhydrous compound remained. what is the formula of the hydrate?
Chemistry
1 answer:
suter [353]3 years ago
8 0
Let the hydrated compound be CaSO4.xH2O
mass of the caso4= 43.63g /(40+32+64+18x)* (40+32+64)= 36.45g
136+18x=( 43.63*136) / 36.45= 162.78
18x = 162.78- 136= 26.8
x = 1.48 = 1.5
Formula of hydrate = CaSO4.1.5H2O = 2CaSo4. 3H2O
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<u>Answer:</u> The formula of limiting reagent is 'CO' and amount of excess reagent remaining is 3.68 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation

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

  • <u>For carbon monoxide:</u>

Given mass of carbon monoxide = 9.16 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon monoxide}=\frac{9.16g}{28g/mol}=0.33mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 9.01 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{9.01g}{32g/mol}=0.28mol

For the given chemical equation:

2CO(g)+O_2(g)\rightarrow 2CO_2(g)

By Stoichiometry of the reaction:

2 mole of carbon monoxide reacts with 1 mole of oxygen gas

So, 0.33 moles of carbon monoxide will react with = \frac{1}{2}\times 0.33=0.165moles of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

So, carbon monoxide is considered as a limiting reagent because it limits the formation of products.

  • Amount of excess reagent (oxygen gas) left = 0.28 - 0.165 = 0.115 moles

Now, calculating the mass of oxygen gas from equation 1, we get:

Moles of oxygen gas = 0.115 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

0.115mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=3.68g

Hence, the formula of limiting reagent is 'CO' and amount of excess reagent remaining is 3.68 grams.

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swat32

Answer:

0.529

Explanation:

Let's consider the reaction A → Products

Since the units of the rate constant are s⁻1, this is a first-order reaction with respect to A.

We can find the concentration of A at a certain time t ([A]_{t}) using the following expression.

[A]_{t}=[A]_{0}.e^{-k\times t}

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