## 1. What volume of gas (NU) is formed by the interaction of 13 g of zinc with 400 g of sulfuric acid solution with a mass fraction of 10%?

Question

1. What volume of gas (NU) is formed by the interaction of 13 g of zinc with 400 g of sulfuric acid solution with a mass fraction of 10%?

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3 years 2021-09-05T12:47:14+00:00 1 Answers 0 views 0

4.5L of gas are produced

Explanation:

Based on the reaction:

Zn + H2SO4 → ZnSO4 + H2(g)

Where 1 mole of Zn reacts with 1 mole of sulfuric acid to produce 1 mole of H2, the gas

To solve this question we must find the moles of each reactant in order to find limiting reactant and, thus, the moles of H2 produced. Then using PV=nRT we can find the volume of the gas:

Moles Zn -Molar mass: 65.38g/mol-

13g * (1mol / 65.38g) = 0.20 moles Zn

Moles H2SO4 -Molar mass: 98g/mol-

400g * 10% = 40g H2SO4 * (1mol / 98g) = 0.41 moles H2SO4

As the ratio of the reaction is 1:1, the limiting reactant is Zn and the moles produced of H2 are 0.20 moles

Using PV = nRT; V = nRT/ P

Where V is the volume of the gas

n are the moles = 0.20 moles

R is gas constant = 0.082atmL/molK

T is absolute temperature = 273.15K at STP

P is pressure = 1atm at STP

V = 0.20mol*0.082atmL/molK*273.15K / 1atm

V = 4.5L of gas are produced