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. __
Answer
4.2
(234 Votes)
Darcy
Professional · Tutor for 6 years
Answer
## The mass of copper carbonate needed is
.## The actual yield of copper chloride is
.
Explanation
## Step1: First, we need to calculate the molar mass of Copper Chloride (CuCl2) and Copper Carbonate (CuCO3). The molar mass of CuCl2 is \(63.5 (Cu) + 2*35.5 (Cl) = 134.5 \, g/mol\). The molar mass of CuCO3 is \(63.5 (Cu) + 12 (C) + 3*16 (O) = 123.5 \, g/mol\).## Step2: 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. ## Step3: 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:###
Solving for x gives us the mass of copper carbonate needed.## Step4: 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%. ###
Given the percentage yield and the theoretical yield, we can calculate the actual yield.#