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2 years ago

In one experiment 7.62 g of Fe are allowed to react with 8.67 g of S

Chemistry

1 answer:

vichka [17]2 years ago

4 0

calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess

moles = mass / molar mass

moles Fe = 7.62 g / 55.85 g/mol

= 0.1364 moles

1 mole Fe produces 1 mole FeS

Therefore 7.62 g Fe can form 0.1364 moles FeS

moles S = 8.67 g / 32.07 g/mol

= 0.2703 moles S

1 mole S can from 1 moles FeS

So 8.67 g S can produce 0.2703 moles FeS

The limiting reagent is the one that produces the least product. So Fe is limiting.

The maximum amount of FeS possible is from complete reaction of all the limiting reagent.

We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.

Convert to mass

hope this helps :)

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As the tempreture of a liquid increases the solubility of a liquid of that liquid

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As the temperature increases, the solubility of the solute in the liquid also increases. This is due to the fact that the increase in energy allows the liquid to more effectively break up the solute. The additoin of energy also shifts the equilibrium of the reation to the right since it takes energy to dissolve most things and you are adding more of it (this is explained with Le Chatlier principles).

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2 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

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