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3 years ago

How many grams of Al2O3 can form from 34.6 grams of Al

Chemistry

1 answer:

ankoles [38]3 years ago

5 0

Answer: 65.28 g of $Al_2O_3$ will be produced from 34.6 g of Al.

Explanation:

To calculate the moles :

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

$\text{Moles of} Al=\frac{34.6g}{27g/mol}=1.28moles$

The balanced chemical reaction is

$4Al+3O_2\rightarrow 2Al_2O_3$

According to stoichiometry :

4 moles of $Al$ produce = 2 moles of $Al_2O_3$

Thus 1.28 moles of $Al$ produce= $\frac{2}{4}\times 1.28=0.64moles$ of $Al_2O_3$

Mass of $[tex]Al_2O_3$ = $moles\times {\text {Molar mass}}=0.64moles\times 102g/mol=65.28g$

Thus 65.28 g of $Al_2O_3$ will be produced from 34.6 g of Al.

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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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Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.100 mole of Ca(N

pickupchik [31]

Answer:

[Ca²⁺] = 1M

[NO₃⁻] = 2M

Explanation:

Calcium nitrate dissociates in water as follows:

Ca(NO₃)₂ ⇒ Ca²⁺ + 2NO₃⁻

The moles of Ca²⁺ can be found using the molar relationship between Ca(NO₃)₂ and Ca²⁺

(0.100mol Ca(NO₃)₂) (Ca²⁺ /Ca(NO₃)₂) = 0.100 mol Ca²⁺

The concentration of Ca²⁺ is then:

[Ca²⁺] = n/V = (0.100mol)/(100.0mL) x (1000ml)/(1L) = 1M

Similarly, moles of NO₃⁻ can be found using the molar relationship between Ca(NO₃)₂ and NO₃⁻:

(0.100mol Ca(NO₃)₂) (2NO₃⁻/Ca(NO₃)₂) = 0.200 mol NO₃⁻

The concentration of NO₃⁻ is then:

[NO₃⁻] = (0.200mol)/(100.0mL) x (1000ml)/(1L) = 2M

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