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Rainbow [258]
1 year ago
10

A student weighs out 0. 0422 g of magnesium metal. The magnesium metal is reacted with excess hydrochloric acid to produce hydro

gen gas. A sample of hydrogen gas is collected over water in a eudiometer at 32. 0°c. The volume of collected gas is 43. 9 ml and the atmospheric pressure is 832 mmhg. Using the experimentally collected data, calculate r and the percent error.
Chemistry
1 answer:
lutik1710 [3]1 year ago
3 0
  • The student weighs out 0.0422 grams of the metal magnesium, thus we can figure that the more's, the magnesium he used, is the mass of the magnesium over the more mass, which is 0.024422.
  • That is approximately 0.001758.
  • Furthermore, it claims that too much hydrochloric acid causes the metal magnesium to react, producing hydrogen gas.
  • The volume of collected gas is 43.9 cc, the mastic pressure is 22 cc, and a sample of hydrogen gas is collected over water in a meter.
<h3>Is it true that calculations made utilizing experimental and gathered data result in a percent error? </h3>
  • Consequently, we are aware that magnesium and chloride react.
  • We create 1 as the reaction ratio is 1:2.
  • The hydrogen and 1 are more.
  • Magnesium chloride is more.
  • Therefore, based on this equation, we can infer that the amount of hydrogen that would be created in this scenario is greater than the amount of magnesium present here, or 0.001758 more.
  • Among hydrogen, there is.
  • \Once we convert the temperature from 32 Celsius to kelvin, we can tell you that the temperature is actually about 5.15 kelvin.
  • The gas has a volume of 43 in m, which is equal to 0.0439 liter and indicates that the pressure is approximately 832 millimeter.
  • Mercury, which is 2 times 13332 plus ca, or roughly 110922.24 par, is a mathematical constant.
  • So, in this instance, we are aware that p v = n r t.
  • The r in this case equals p v over n t, thus we want to determine the r.
  • So p is 110922.24. The temperature is 305.15 and the V is 0.04  over the n is 0.001758.
  • Let's proceed with the calculations right now.
  • In this instance, you will discover that the solution is 9.077 times 10; that is all there is to it.

To learn more about Magnesium chloride reactions visit:

brainly.com/question/27891157

#SPJ4

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\large \boxed{\text{-41.2 kJ/mol}}

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We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)

(a) Enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}

(b) Total enthalpies of reactants and products

\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

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To proceed, we have to obtain the limiting reagent,

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