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(c) Gold reacts with the elements in Group 7 of the periodic table. 0.175 g of gold reacts with chlorine. The equation for the reaction is: 2Au+3Cl_(2)arrow 2AuCl_(3) Calculate the mass of chlorine needed to react with 0.175 g of gold. Give your answer in mg Relative atomic masses (A); Cl=35.5Au=197

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(c) Gold reacts with the elements in Group 7 of the periodic table.
0.175 g of gold reacts with chlorine.
The equation for the reaction is:
2Au+3Cl_(2)arrow 2AuCl_(3)
Calculate the mass of chlorine needed to react with 0.175 g of gold.
Give your answer in mg
Relative atomic masses (A); Cl=35.5Au=197

(c) Gold reacts with the elements in Group 7 of the periodic table. 0.175 g of gold reacts with chlorine. The equation for the reaction is: 2Au+3Cl_(2)arrow 2AuCl_(3) Calculate the mass of chlorine needed to react with 0.175 g of gold. Give your answer in mg Relative atomic masses (A); Cl=35.5Au=197

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ZaidProfessional · Tutor for 6 years

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<br /><br />Step 1: Find the molar masses of gold (Au) and chlorine gas (Cl2).<br /><br />Given the relative atomic masses: Au = 197 and Cl = 35.5 <br />However, the reaction equation involves Cl2, a chlorine molecule which contains two Cl atoms. Therefore, it's necessary to calculate the molar mass of chlorine molecule as: <br /><br />\[ Mass(Cl2) = 2* Mass(Cl) = 2*35.5g/mol =71g/mol \] <br /><br />and the molar mass of \(Au=A_{r}Au = 197 g/mol \)<br /><br />Step 2: Identify the stoichiometry of the reaction.<br /><br />According to the balanced equation, This equation says that 2 moles of \(Au\) react with 3 moles of \(Cl2\) to form 2 moles of \(AuCl3\). Therefore, the ratio of the moles of \(Au\) to \(Cl2\) is \(2:3\).<br /><br />Step 3: Calculate the number of moles in 0.175g of \(Au\).<br /><br />\[<br />Moles = mass/molar\ mass\\ <br />Moles = 0.175g/197g/mol = 9 * 10^{-4}mol.<br />\] <br /><br />Step 4: Determine the number of moles of \(Cl_2\) involved in the 9 * \(10^{-4}\) mol of \(Au\). <br />Due to the 2 to 3 stoichiometric ratio: mol(\(Cl_2\)) = \(\frac{3}{2} * \) mol(\(Au\))<br /><br />\[<br /> mol(Cl_{2}) = \frac{3}{2} * 9 * 10^{-4} mol(Yttruium) = 1.35 * 10^{-3} mol<br />\]<br /><br />Step 5: Convert this into a mass to find the mass of chlorine needed. <br /><br />\[ <br />Mass = moles * molar \ mass \\<br />Mass = 1.35 * 10^{-3} mol * 71 g/mol = 0.096 g of Cl2 \]. <br /><br />Because the answer needs to be in milligrams, we just multiply by 1000 to convert. <br />So <br />\[ <br />Mass = 0.096 g * 1000 = 96 mg.<br />\]<br /><br />Hence, **96 mg of Cl2 is the final result** needed to react with 0.175 g of Au according to the balanced equation.pdf
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