Question

A gas is formed in the reaction shown below. the gas laws can help determine the volume of gas produced.

Zn(s) + 2 HCl(aq) → ZnCl₂ + H₂(g)

you have determined that 1.961 moles of h₂(g) were produced in this reaction. at stp, what volume in l of H₂ will be produced? remember the gas constant, r, is 0.08206 l･atm/mol･k.

Answer 1

The volume of H2 gas produced at STP by 1.961 moles is 44.15 liters, determined by the ideal gas law and the known molar volume at STP.

Step-by-step explanation:

At STP, 1.961 moles of H2 gas will occupy a volume of 44.15 liters, since 1 mole of gas at STP occupies 22.4 liters.

Using the ideal gas law, PV = nRT, we can calculate the volume of a gas when the number of moles (n), the temperature (T), and the pressure (P) are known. At STP, where P = 1 atm and T = 273.15 K, the volume (V) can be directly found for any amount of an ideal gas by using the molar volume of 22.4 L/mol. When 1.961 moles of H2 are produced in the reaction Zn(s) + 2 HCl(aq) → ZnCl2 (aq) + H2(g), the volume of hydrogen gas can be calculated by multiplying the number of moles by the molar volume: V = 1.961 moles * 22.4 L/mol.