Question

question 7
given the balanced equation below, how many grams of sodium chloride will be produced from 1.87 grams of copper (ii) chloride?
round your answer to two decimal places and do not include units.
2 na3po4 (aq) + 3 cucl2 (aq) → cu3(po4)2 (s) + 6 nacl (aq)
1 point
page 7 of 8
1.63

Explanation:

Step1: Calculate the molar mass of \(CuCl_2\)

The molar mass of \(Cu\) is \(63.55\space g/mol\), and the molar mass of \(Cl\) is \(35.45\space g/mol\).
For \(CuCl_2\), \(M = 63.55+(2\times35.45)=63.55 + 70.9=134.45\space g/mol\)

Step2: Calculate the number of moles of \(CuCl_2\)

Given \(m = 1.87\space g\), using \(n=\frac{m}{M}\), we have \(n=\frac{1.87}{134.45}\approx0.0139\space mol\)

Step3: Use the stoichiometry of the reaction

From the balanced equation \(2Na_3PO_4(aq)+3CuCl_2(aq)\to Cu_3(PO_4)_2(s)+6NaCl(aq)\), the mole ratio of \(CuCl_2\) to \(NaCl\) is \(3:6 = 1:2\)
So the number of moles of \(NaCl\) is \(n_{NaCl}=2\times n_{CuCl_2}=2\times0.0139 = 0.0278\space mol\)

Step4: Calculate the molar mass of \(NaCl\)

The molar mass of \(Na\) is \(22.99\space g/mol\), and the molar mass of \(Cl\) is \(35.45\space g/mol\).
For \(NaCl\), \(M = 22.99+35.45 = 58.44\space g/mol\)

Step5: Calculate the mass of \(NaCl\)

Using \(m = n\times M\), we have \(m=0.0278\times58.44\approx1.62\)

Answer:

\(1.62\)