Question

strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown. srcl₂(aq) + 2naf(aq) → srf₂(s) + 2nacl(aq) what volume of a 0.670 m naf solution is required to react completely with 783 ml of a 0.180 m srcl₂ solution? volume: ml how many moles of srf₂ are formed from this reaction? moles of srf₂: mol

Explanation:

Step1: Calculate moles of \(SrCl_2\)

Use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume in liters).
\(V_{SrCl_2}=783\space mL = 0.783\space L\), \(C_{SrCl_2}=0.180\space M\)
\(n_{SrCl_2}=0.180\times0.783 = 0.14094\space mol\)

Step2: Find moles of \(NaF\) using mole - ratio

From the balanced equation \(SrCl_2(aq)+2NaF(aq)\to SrF_2(s)+2NaCl(aq)\), the mole - ratio of \(SrCl_2:NaF = 1:2\)
\(n_{NaF}=2\times n_{SrCl_2}\)
\(n_{NaF}=2\times0.14094 = 0.28188\space mol\)

Step3: Calculate volume of \(NaF\) solution

Use \(V=\frac{n}{C}\), \(C_{NaF} = 0.670\space M\)
\(V_{NaF}=\frac{0.28188}{0.670}=0.4207\space L=421\space mL\)

Step4: Calculate moles of \(SrF_2\)

From the mole - ratio \(SrCl_2:SrF_2=1:1\)
\(n_{SrF_2}=n_{SrCl_2}=0.14094\approx0.141\space mol\)

Answer:

volume: \(421\space mL\)
moles of \(SrF_2\): \(0.141\space mol\)