Question

ideal gas law pv = nrt
universal gas constant r = 0.0821 l·atm/(mol·k)
standard atmospheric pressure 1 atm = 101.3 kpa
celsius to kelvin conversion k = °c + 273.15

1
1 h 1.008

select the correct answer.
when magnesium reacts with hydrochloric acid, hydrogen gas is formed: mg(s)+2hcl(aq)=h₂(g)+mgcl₂(aq)
what 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.
a. 1.98 l
b. 2.96 l
c. 48.4 l
d. 92.9 l
e. 99.1 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.

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 \(n_{HCl}=\frac{46.5\ g}{36.458\ g/mol}\approx1.275\ mol\).

1. 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 \(n_{H_2}=\frac{1}{2}\times1.275\ mol = 0.6375\ mol\).

1. 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. \(15.8\ L\)