Question

Read the given equation.
2Na+ 2H₂O 2NaOH + H₂
During a laboratory experiment, a certain quantity of sodium metal reacted with water to produce sodium hydroxide and hydrogen gas. What was the initial quantity of
sodium metal used if 7.80 liters of H₂ gas were produced at STP?
07:29 grams
09.30 grams
12.2 grams
16.0 grams

Answer 1

Answer:

From the equation it is evident that 2 moles of Sodium metal produces 1 mole of hydrogen gas.

At STP 1mole of any gas occupies a volume of 22.4 litres.

Therefore, 7.80 gives---(7.80x1)/22.4 moles = 0.3482 moles

Since the mole ratio of Sodium to hydrogen is 2:1, then the number of moles of sodium that reacted is given by the following expression.

(0.3482 * 2) / 1 moles which gives 0.6964 moles.

The atomic mass of sodium is 23 thus the mass of sodium that reacted is given by:

mass=no. of Monogram

0.6964 * 23 = 16.02 grams.

Explanation:

please mark me brainlist