Question

Consider the reaction between copper(II) nitrate and strontium chloride to yield copper(II) chloride and strontium nitrate. How many grams of strontium chloride would be needed to completely react 5.184 grams of copper(II) nitrate?

Answer 1

Cu(NO₃)₂ + SrCl₂ → CuCl₂ + Sr(NO₃)₂

M(Cu(NO₃)₂) = 187.56 g/mol

n(Cu(NO₃)₂) $= \frac{5.184}{187.56} = 0.0276 \;mol$

n(SrCl₂) = n(Cu(NO₃)₂) = 0.0276 mol

M(SrCl₂) = 158.53 g/mol

m(SrCl₂) $= 0.0276 \times 158.53 = 4.38 \;g$