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3 years ago

How much heat is released when 24.8 g of ch4 is burned in excess oxygen gas?

Chemistry

1 answer:

balu736 [363]3 years ago

4 0

The given question is incomplete. The complete question is:

How much heat is produced when 24.8 g of $CH_4$ is burned in excess oxygen gas

Given: $CH _4
+2O_2\rightarrow CO_2+2H_2O$ ΔH= −802 kJ.

Answer: 1243.1 kJ

Explanation:

Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.

Given :

Amount of heat released on combustion of 1 mole of methane = 802 kJ kJ/mol

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of $CH_4$ weighs = 16 g

Thus we can say:

16 g of $CH_4$ on combustion releases heat = 802 kJ

Thus 24.8 g of $CH_4$ on combustion releases = $\frac{802}{16}\times 24.8=1243.1kJ$

Thus heat released when 24.8 g of methane is burned in excess oxygen gas is 1243.1 kJ

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Answer the following questions based on the reaction below: NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)

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b) 77.3%

Explanation:

1. Chemical balanced equation (given)

$NaOH(aq)+ KHP(s)\rightarrow NaKP(aq)+H_2O(l)$

2. Mole ratio

$1molNaOH(aq):1molKHP(s)$

This is, 1 mol of NaOH will reacts with 1 mol of KHP.

3. Find the number of moles in 72.14 mL of the base

$Molarity=\text{number of moles of solute}/\text{volume of solution in liters}$

$\text{Volume of solution}=72.14mL=0.07214liters$

$\text{Number of moles of NaOH}=0.0978M\times 0.07214liter=0.007055mol$

4. Find the number of grams of KHP that reacted

The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol

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mass = number moles × molar mass = 0.007055mol × 204.23g/mol

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You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).

5. Find the percentage of KHP in the sample

The percentage is how much of the substance is in 100 parts of the sample.

The formula is:

% = (mass of substance / mass of sample) × 100

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