How many liters of chlorine gas can react with 56.0 grams of calcium metal at standard temperature and pressure? Show all of the work used to find your answer.Ca + Cl2 CaCl2

Answers

Answer 1

Answer: The balanced chemical reaction is:

Ca + Cl2 = CaCl2

We are given the amount of calcium metal to be used for this reaction. This will be the starting point for the calculations.

56 g Ca ( 1 mol Ca / 40.08 g Ca) (1 mol Cl2 / 1 mol Ca) ( 22.414 L Cl2 / 1 mol Cl2 ) = 31.32 L Cl2 gas produced from the reaction

Answer 2

Answer:

Answer:

31.36 L of chlorine gas will react with 56 g of calcium.

Explanation:

$Ca + Cl_2\rightarrow CaCl_2$

Mass of calcium = 56 g

Moles of calcium gas = $(56 g)/(40 g/mol)=1.4 moles$

According to reaction, 1 mole of calcium react with 1 mole of chlorine gas.

Then 1.4 mol of calcium wiull react with :

$(1)/(1)* 1.4 = 1.4 mol$ chlorine gas

At standard temperature and pressure. 1 mol of gas occupies 22.4 L of volume.

Then 4.1 moles of chlorine gas will occupy :

$22.4 L\tmes 4.1 =31.36 L$

31.36 L of chlorine gas will react with 56 g of calcium.

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Answers

Answer:

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Explanation:

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Answers

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Answers

Given:

64.82 g of aluminum metal

100.0oC is cooled to 82.0 °C

specific heat of aluminum is 0.897 J /g °C.

Required:

Change in heat energy

Solution:

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H = mCpT

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Answers

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Answers

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Answers

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