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4 years ago

If 27.9 grams of calcium carbonate reacts with excess HCl, how many moles of calcium chloride are produced

Chemistry

1 answer:

Virty [35]4 years ago

5 0

Answer:

$\large \boxed{\text{0.2797 mol}}$

Explanation:

We must do the conversions:

mass of CaCO₃ ⟶ moles of CaCO₃ ⟶ moles of CaCl₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 100.09 110.98

CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + 2HCl

m/g: 27.9

(a) Moles of CaCO₃ $\text{Moles of CaCO}_{3} = \text{27.9 g CaCO}_{3}\times \dfrac{\text{1 mol CaCO}_{3}}{\text{100.09 g CaCO}_{3}}= \text{0.2787 mol CaCO}_{3}$

(b) Moles of CaCl₂ $\text{Moles of CaCl}_{2} =\text{0.2787 mol CaCO}_{3} \times \dfrac{\text{1mol CaCl}_{2}}{\text{1 mol CaCO}_{3}} = \text{0.279 mol CaCl}_{2}$ $\text{The reaction produces $\large \boxed{\textbf{0.279 mol}}$ of CaCl}_{2}$

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

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3 years ago

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Assoli18 [71]

Answer:

Sig fig 3 decimal 6

Explanation:

Hii

......

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