Can I please get the answer to this question???

Question

How many liters of hydrogen gas at STP are produced from the single displacement reaction of 52.3 g of zinc and excess sulfuric acid?

Zn  + H2SO4   →   ZnSO4   + H2

_____unit:_______

Answer 1

Convert 52.3 g of zinc into moles (because the chemical reaction equation is in units of moles).

Look up zinc in the Periodic Table and find it is 65.382 g/mol so:

(52.3 g)/(65.382 g/mol0 = 0.7999 moles of Zn.

In looking at the chemical reaction equation, we see that for every 1 mole of Zn, 1 mole of H₂ is created so there are 0.7999 moles of H₂ created.

The ideal gas law can be used now to convert H₂ into liters of H₂.

The Ideal Gas Law says PV = nRT or, since we are looking for volume (liters) then V = nRT/P

n = 0.7999 moles

At STP, P = 1 atm and T = 0 °C or 273.15 K

R (a constant we look up) is 0.082057 L⋅atm/(K⋅mol)

So V = nRT/P

V = (0.7999)(0.082057)(273.15)/1 = 17.929 Liters

Since there are 3 sig figs used in the givens, round the answer to 3 sig figs to get V = 17.9 L

Answer 2

*ideal gas law: gases under STP all gases have a volume of 22.4 L/mol*

1. We are given the amount of one of the reactants, 52.3 g of zinc
2. From the periodic table: the molar mass of Zn is 65.39 grams per mole

52.3 g Zn x 1 mol Zn = 0.7998 mol Zn

65.39 g

Based on our equation;

Zn  + H₂SO₄   →   ZnSO₄   + H₂

we know that for every one mol of Zn reactant, we yield one mol of H₂, so there will also be 0.7998 mol H₂

using the ideal gas law:

0.7998 mol H₂ x 22.4 L = 17.92 L H₂ @ STP

1 mol