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?

 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 liters.

Therefore, 7.80 gives---(7.80×1)/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 moles×RAM

0.6964×23= 16.02 grams.

answer: i would say b.

explanation:

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