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kolbaska11 [484]
1 year ago
7

Zinc metal (Zn) will react with aqueous hydrochloric acid (HCI aq) to produce aqueous zinc chloride (ZnCl2 aq) and hydrogen gas

(H2). Which of the following isthe complete, balanced equation for this reaction?1. 2 HCI (aq) + 2 Zn (s) > 2 H2 (g) + ZnCl2 (aq)2. HCI (aq) + Zn (s) > H2 (g) + ZnCl2 (ag)3. 2 HCI (aq) + Zn (s) › 2 H2 (g) + ZnCl2 (ag)4. 2 HCI (aq) + Zn (s) › H2 (g) + 2 ZnCl2 (ag)5. 2 HCI (aq) + Zn (s) › H2 (g) + ZnCl2 (ad)
Chemistry
1 answer:
Debora [2.8K]1 year ago
3 0

Answer:

5

Explanation:

Here, we want to get the equation of the reaction between Hydrochloric acid and Zinc metal

Zinc metal displaces the hydrogen from hydrochloric acid to form zinc chloride

We have the equation of reaction as:

2HCl_{(aq)}\text{ + Zn}_{(s)}\text{ }\rightarrow\text{ ZnCl}_{2(aq)}\text{ + H}_{2(g)}

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How many litres (L) of oxygen at STP can be obtained from 110 g of potassium chlorate? [R=0.0821 L.atm/K.mol]
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Balance the equation first:

2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)

Moles of KClO3 =  110 / 122.5 = 0.89

Following the balanced chemical equation:
We can say moles of O2 produce =  \frac{3}{2} x moles of KClO3

So, O2 = (3 / 2) x  0.89

= 1.34 moles

So, Volume at STP = nRT / P

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3 0
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An aqueous solution of a soluble compound (a nonelectrolyte) is prepared by dissolving 33.2 g of the compound in sufficient wate
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Answer:

2710.2g/mol

Explanation:

Step 1:

Data obtained from the question. This include the following:

van 't Hoff factor (i) = 1 (since the compound is non-electrolyte)

Mass of compound = 33.2g

Volume = 250mL

Osmotic pressure (Π) = 1.2 atm

Temperature (T) = 25ºC = 25ºC + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Molar mass of compound =.?

Step 2:

Determination of the molarity of the compound.

The molarity, M of the compound can be obtained as follow:

Π = iMRT

1.2 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 1.2 / (0.0821 x 298)

M = 0.049mol/L

Step 3:

Determination of the number of mole of compound in the solution. This can be obtain as follow:

Molarity = 0.049mol/L

Volume = 250mL = 250/1000 = 0.25L

Mole of compound =..?

Molarity = mole /Volume

0.049 = mole / 0.25

Cross multiply

Mole = 0.049 x 0.25

Mole of compound = 0.01225 mole.

Step 4:

Determination of the molar mass of the compound. This is illustrated below:

Mole of the compound = 0.01225 mole.

Mass of the compound = 33.2g

Molar mass of the compound =.?

Mole = Mass /Molar Mass

0.01225 = 33.2/Molar Mass

Cross multiply

0.01225 x molar mass = 33.2

Divide both side by 0.01225

Molar mass = 33.2/0.01225

Molar mass of the compound = 2710.2g/mol

Therefore, the molar mass of the compound is 2710.2g/mol

6 0
3 years ago
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