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3 years ago

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How many grams of zinc metal will react completely with 5.4 liters of 2.8 M HCl? Show all of the work needed to solve this probl

em. Zn (s) + 2 HCl (aq) yields ZnCl2 (aq) + H2 (g)

2 answers:

Assoli18 [71]3 years ago

5 0

The reaction is
<span>Zn (s) + 2 HCl (aq) ----> ZnCl2 (aq) + H2 (g)
which is already balanced
5.4 L of 2.8 M HCl contains
5.4 L (2.8 M) = 15.12 moles HCl
The amount of Zinc that will react completely with the acid is
15.12 mol HCl (1 mol Zn / 2 mol HCl) (65 g Zn/1 mol Zn) = 491.4 g Zn

</span>

creativ13 [48]3 years ago

4 0

Answer : The amount of zinc metal react will be, 494.27 grams.

Solution : Given,

Volume of HCl solution = 5.4 L

Molarity of HCl solution = 2.8 M

Molar mass of zinc (Zn) = 65.38 g/mole

First we have to calculate the moles of HCl.

$\text{Molarity of }HCl=\frac{\text{Moles of }HCl}{\text{Volume of solution}}$

$2.8M=\frac{\text{Moles of }HCl}{5.4L}$

$\text{Moles of }HCl=15.12moles$

Now we have to calculate the moles of Zn.

The given balanced chemical reaction is,

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

From the balanced reaction, we conclude that

As, 2 moles of $HCl$ react to with 1 mole of Zn

So, 15.12 moles of $HCl$ react to with $\frac{15.12}{2}=7.56$ moles of Zn

Now we have to calculate the mass of Zn.

$\text{Mass of }Zn=\text{Moles of }Zn\times \text{Molar mass of }Zn$

$\text{Mass of }Zn=(7.56mole)\times (65.38g/mole)=494.27g$

Therefore, the amount of zinc metal react will be, 494.27 grams.

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If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated

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Answer : The molar mass of the solute would be low.

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$w_b$ = mass of solute

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$M_b$ = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

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4 years ago

P4O10 -> 4P+5O2 How many moles of phosphorus would be produced if 5.3 mol of P4O10 reacted?

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3 years ago

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

sp2606 [1]

Answer:

Mass of CaCl₂ = 20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted = 7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

CaCO₃ : CaCl₂

1 : 1

0.25 : 0.25

HCl : CaCl₂

2 : 1

0.36 : 1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ = 20 g

The calcium carbonate is present in excess.

HCl : CaCO₃

2 : 1

0.36 : 1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted = 7.007 g

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3 years ago