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(b) A student wanted to make 11.0 g of copper chloride. The equation for the reaction is: CuCO_(3)+2HClarrow CuCl_(2)+H_(2)O+CO_(2) Relative atomic masses, A:H=1;C=12;O=16;Cl=35.5;Cu=63.5 Calculate the mass of copper carbonate the student should react with dilute hydrochloric acid to make 11.0 g of copper chloride. __ Mass of copper carbonate= __ (c) The percentage yield of copper chloride was 79.1% Calculate the mass of copper chloride the student actually produced. __

Question

(b) A student wanted to make 11.0 g of copper chloride.
The equation for the reaction is:
CuCO_(3)+2HClarrow CuCl_(2)+H_(2)O+CO_(2)
Relative atomic masses, A:H=1;C=12;O=16;Cl=35.5;Cu=63.5
Calculate the mass of copper carbonate the student should react with dilute
hydrochloric acid to make 11.0 g of copper chloride.
__
Mass of copper carbonate=
__
(c) The percentage yield of copper chloride was 79.1% 
Calculate the mass of copper chloride the student actually produced.
__

(b) A student wanted to make 11.0 g of copper chloride. The equation for the reaction is: CuCO_(3)+2HClarrow CuCl_(2)+H_(2)O+CO_(2) Relative atomic masses, A:H=1;C=12;O=16;Cl=35.5;Cu=63.5 Calculate the mass of copper carbonate the student should react with dilute hydrochloric acid to make 11.0 g of copper chloride. __ Mass of copper carbonate= __ (c) The percentage yield of copper chloride was 79.1% Calculate the mass of copper chloride the student actually produced. __

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DarcyProfessional · Tutor for 6 years

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## The mass of copper carbonate needed is \( x = \frac{123.5 \, g \, CuCO3 \times 11.0 \, g \, CuCl2}{134.5 \, g \, CuCl2} = 10.1 \, g \, CuCO3 \).<br /><br />## The actual yield of copper chloride is \( \text{Actual yield} = \frac{\text{Percentage yield} \times \text{Theoretical yield}}{100\%} = \frac{79.1\% \times 11.0 \, g}{100\%} = 8.7 \, g \, CuCl2 \).

Explain

## Step1: <br />First, we need to calculate the molar mass of Copper Chloride (CuCl2) and Copper Carbonate (CuCO3). <br /><br />The molar mass of CuCl2 is \(63.5 (Cu) + 2*35.5 (Cl) = 134.5 \, g/mol\). <br /><br />The molar mass of CuCO3 is \(63.5 (Cu) + 12 (C) + 3*16 (O) = 123.5 \, g/mol\).<br /><br />## Step2: <br />From the balanced chemical equation, we know that 1 mole of Copper Carbonate (CuCO3) reacts to produce 1 mole of Copper Chloride (CuCl2). Because the reaction is one-to-one, the measurement of CuCO3 and CuCl2 in moles is equivalent. <br /><br />## Step3: <br />To find out how much copper carbonate is needed to produce 11.0 g of copper chloride, we can set up a proportion using the molar masses:<br /><br />### \( \frac{123.5 \, g \, CuCO3}{134.5 \, g \, CuCl2} = \frac{x \, g \, CuCO3}{11.0 \, g \, CuCl2} \)<br /><br />Solving for x gives us the mass of copper carbonate needed.<br /><br />## Step4: <br />The percentage yield of copper chloride was 79.1%. The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%. <br /><br />### \( \text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\% \)<br /><br />Given the percentage yield and the theoretical yield, we can calculate the actual yield.<br /><br />#
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