ideal gas law pv = nrt\nuniversal gas constant r = 0.0821 l·atm/(mol·k)\nstandard atmospheric pressure 1 atm…

ideal gas law pv = nrt\nuniversal gas constant r = 0.0821 l·atm/(mol·k)\nstandard atmospheric pressure 1 atm = 101.3 kpa\ncelsius to kelvin conversion k = °c + 273.15\n\n1\n1 h 1.008\n\nselect the correct answer.\nwhen magnesium reacts with hydrochloric acid, hydrogen gas is formed: mg(s)+2hcl(aq)=h₂(g)+mgcl₂(aq)\nwhat is the volume of hydrogen produced at 27°c and 90.3 kpa when 48.6 grams of mg reacts with excess hydrochloric acid? use the given table of ideal - gas constants.\na. 1.98 l\nb. 2.96 l\nc. 48.4 l\nd. 92.9 l\ne. 99.1 l

ideal gas law pv = nrt\nuniversal gas constant r = 0.0821 l·atm/(mol·k)\nstandard atmospheric pressure 1 atm = 101.3 kpa\ncelsius to kelvin conversion k = °c + 273.15\n\n1\n1 h 1.008\n\nselect the correct answer.\nwhen magnesium reacts with hydrochloric acid, hydrogen gas is formed: mg(s)+2hcl(aq)=h₂(g)+mgcl₂(aq)\nwhat is the volume of hydrogen produced at 27°c and 90.3 kpa when 48.6 grams of mg reacts with excess hydrochloric acid? use the given table of ideal - gas constants.\na. 1.98 l\nb. 2.96 l\nc. 48.4 l\nd. 92.9 l\ne. 99.1 l

Answer

Answer:

  1. First, find the molar mass of (HCl). The molar mass of (H = 1.008\ g/mol) and (Cl=35.45\ g/mol), so (M_{HCl}=1.008 + 35.45=36.458\ g/mol).
    • The number of moles of (HCl), (n=\frac{m}{M}), where (m = 46.5\ g). So (n_{HCl}=\frac{46.5\ g}{36.458\ g/mol}\approx1.275\ mol).
  2. Then, use the stoichiometry of the reaction (2HCl+Mg = H_2+MgCl_2).
    • From the balanced - chemical equation, the mole ratio of (HCl) to (H_2) is (2:1). So the number of moles of (H_2) produced, (n_{H_2}=\frac{1}{2}n_{HCl}).
    • Substituting (n_{HCl}=1.275\ mol), we get (n_{H_2}=\frac{1}{2}\times1.275\ mol = 0.6375\ mol).
  3. Next, use the ideal - gas law (PV = nRT).
    • The temperature (T=(27 + 273.15)\ K=300.15\ K) (converting (27^{\circ}C) to Kelvin), (P = 99.3\ kPa=99300\ Pa), and (R = 8.314\ J/(mol\cdot K)).
    • We want to find the volume (V), and from (PV = nRT), we can solve for (V): (V=\frac{nRT}{P}).
    • Substituting (n = 0.6375\ mol), (R = 8.314\ J/(mol\cdot K)), (T = 300.15\ K), and (P = 99300\ Pa) into the formula:
      • (V=\frac{0.6375\ mol\times8.314\ J/(mol\cdot K)\times300.15\ K}{99300\ Pa}).
      • First, calculate the numerator: (0.6375\times8.314\times300.15\approx1590.7\ J).
      • Then, (V=\frac{1590.7\ J}{99300\ Pa}\approx0.016\ m^3).
      • Since (1\ m^3 = 1000\ L), (V = 16\ L).

The closest answer among the options (assuming some rounding differences in the options) is: A. (15.8\ L)

Explanation:

Step1: Calculate moles of (HCl)

  • Use (n=\frac{m}{M}), with (m = 46.5\ g) and (M = 36.458\ g/mol).

Step2: Determine moles of (H_2)

  • Use mole - ratio from balanced equation ((HCl:H_2 = 2:1)).

Step3: Apply ideal - gas law

  • Rearrange (PV = nRT) to (V=\frac{nRT}{P}) and substitute values.