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3 years ago

How much energy is lost to condense 300. grams of steam at 100.C?

Chemistry

1 answer:

True [87]3 years ago

3 0

Energy lost to condense = 803.4 kJ

<h3>Further explanation</h3>

Condensation of steam through 2 stages:

1. phase change(steam to water)

2. cool down(100 to 0 C)

1. phase change(condensation)

Lv==latent heat of vaporization for water=2260 J/g

$\tt Q=300\times 2260=678000~J$

2. cool down

c=specific heat for water=4.18 J/g C

$\tt Q=300\times 4.18\times (100-0)=125400$

Total heat =

$\tt 678000+125400=803400~J$

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Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

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Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

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Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

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Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

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On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

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