# what volume of hydrogen is evolved when 20 g of zinc is reacted with excess dilute sulphuric VI acid at STP

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what volume of hydrogen is evolved when 20 g of zinc is reacted with excess dilute sulphuric VI acid at STP

Molar mass of Zn = 65.38 g/mol
The balanced equation:
Zn(s) + H2SO4(aq)
→ ZnSO4(aq) + H2(g)

1. Moles of Zn in 20.0 g = (mass of Zn in g/molar mass of Zn in g/mol) = (20.0 g/65.38 g/mol) = 0.3059 mol
2. From the balanced equation, the mole ratio of Zn to H2 is 1:1.
3. Therefore, for 0.3059 mol of Zn, moles of H2 produced in the reaction = (1/1) x 0.3059 mol = 0.3059 mol
4. At standard temperature and pressure (STP), the pressure is 1 bar (100 kPa or 0.986923 atm) and the temperature is 0 deg C (273.15 K), 1 mole of an ideal gas occupies a volume of 22.71 L
5. Therefore, for 0.3059 mol of H2, the volume of gas produced in the reaction = (0.3059 mol x 22.71 L/mol) = 6.95 L or approx. 7.0 L
6. Thus, the volume of hydrogen evolved when 20.0 g of zinc is reacted with excess dilute sulphuric acid at STP = 7.0 L