How many liters of hydrogen are obtained from the reaction of 4.00 g sodium with excess water, at STP?

Question

How many liters of hydrogen are obtained from the reaction of 4.00 g sodium with excess water, at STP?

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Khánh Gia 5 months 2021-08-15T13:18:27+00:00 1 Answers 2 views 0

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    2021-08-15T13:20:11+00:00

    Answer:

    V = 1.95 L.

    Explanation:

    Hell there!

    In this case, according to the following reaction between sodium metal and water:

    2Na+2H_2O\rightarrow 2NaOH+H_2

    We can realize that the moles of hydrogen can be calculated by using the initial mass of sodium, its atomic mass (23.0 g/mol) and the 2:1 mole ratio of sodium to hydrogen to obtain:

    4.00gNa*\frac{1molNa}{23.0gNa} *\frac{1molH_2}{2molNa}=0.0870molH_2

    Finally, we calculate the volume of hydrogen by using the ideal gas equation whereas the pressure is 1 atm and the temperature 273.15 K according to the STP conditions:

    PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0870mol*0.08206\frac{atm*L}{mol*K}*273.15 K}{1 atm}\\\\V=1.95L

    Regards!

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