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goldfiish [28.3K]
3 years ago
14

In testing the effectiveness of an antacid compound, 20.0g of Hydrochloric Acid is mixed with 28.0g of Magnesium Hydroxide. Will

the base neutralize (Completely use up) all of the acid? How much of which substance is in excess? 2HCL + (OH)2 -> MgCl2 + 2H20
Chemistry
1 answer:
BartSMP [9]3 years ago
4 0

Answer:

Base Mg(OH)2 does neutralise the acid and is 12g in excess.

Explanation:

2HCL +Mg(OH)2 -> MgCl2 + 2H20

2 * 36.458 g of HCL react with  58.319 g of  Mg(OH)2 to neutralise it.

72.916 HCl reacts with  58.319 g of the base.

So 20 g HCl reacts with  (58.319/72.916) * 20 = 16g.

There are 28 g of Mg(OH)2 so the base does neutralise all the acid.

The Mg(OH)2 is 28 - 16 = 12 g in excess.

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What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

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Answer:

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Explanation:

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3 years ago
An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its
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Hello!

First you need to calculate q 
<span>delta U is change in internal energy </span>

<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>

<span>delta U = q + w </span>

<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>

<span>q = m x c x delta T </span>

<span>7211 J = 80.0 g x c x (225-25) °C </span>

<span>c = 0.451 J /g °C
</span>
Hope this Helps! Have A Wonderful Day! :)
5 0
3 years ago
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