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Find the mass of sodium required to reduce 6.58 L of hydrogen gas at 32°C and 895 mm, when sodium reacts with hydrochloric acid. (Hint: you first need to write a balanced chemical equation.)

Find the mass of sodium required to reduce 658 L of hydrogen gas at 32C and 895 mm when sodium reacts with hydrochloric acid Hint you first need to write a bala class=

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Answer:

Mass = 14.72 g

Explanation:

Given data:

Volume of hydrogen = 6.58 L

Temperature of gas = 32°C (32+273 = 305 k)

Pressure of gas = 895 mmHg (895/760 = 1.2 atm)

Mass of sodium required = ?

Solution:

Chemical equation:

2HCl + 2Na     →      2NaCl + H₂

Number of moles of hydrogen:

PV = nRT

1.2 atm × 6.58 L = n× 0.0821 atm.L/mol.K×305 K

7.9 atm.L =  n×  25.0  atm.L/mol

n = 7.9 atm.L / 25.0  atm.L/mol

n = 0.32 mol

Now we will compare the moles of hydrogen with sodium.

             H₂           :        Na

              1              :        2

            0.32          :     2/1×0.32 = 0.64 mol

Mass of sodium:

Mass = number of moles × molar mass

Mass = 0.64 mol × 23 g/mol

Mass = 14.72 g

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