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 8.40 liters of h2 gas were produced at stp? a. 17.2 grams b. 15.8 grams c. 12.3 grams d. 10.0 grams

                17.25 g of Na

                        The balance chemical equation is as follow,

                              2 Na + 2 H₂O    →    2 NaOH + H₂

As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ (1 L) volume.

Then, according to equation,

            22.4 L (1 mol) H₂ was produced by  =  46 g (2 mol) of Na

Then,

                     8.40 L H₂ gas will be produced by  =  X g of Na

Solving for X,

                      X =  (8.40 L × 46 g) ÷ 22.4 L

                      X =  17.25 g of Na

A) 17.2 grams

Explanation:

I took the quiz and A was correct!

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.