Question

Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown. SrCl2(am)+2NaF(aq) —-> SrF2(s)+2NaCl(aq). What volume of a 0.510 M NaF solution is required to react completely with 439 mL of a 0.660 M SrCl2 solution

Answer 1

Answer:

1,14L of 0.510M NaF solution

Explanation:

For the reaction:

SrCl₂(aq) + 2NaF(aq) → SrF₂(s) + 2NaCl(aq)

Moles in 439mL, 0,439L of a 0.660M SrCl₂ are:

0,439L×0,660mol/L = 0,290 moles of SrCl₂

As 1 mole of SrCl₂ reacts with 2 moles of NaF:

0,290 moles of SrCl₂×(2mol NaF / 1mol SrCl₂) = 0,580mol NaF.

These moles of NaF are required for a complete reaction with SrCl₂. In a 0.510M NaF solution you have 0.580mol in:

0,580mol × (1L/0,510mol) = 1,14L of 0.510M NaF solution

I hope it helps!

Answer 2

The volume of the NaF solution required to react completely with the given SrCl₂ is 1136 mL

From the question,

We are to determine the volume of 0.510 M NaF solution is required to react completely with 439 mL of a 0.660 M SrCl₂ solution

The given balanced chemical equation for the reaction is

SrCl₂(aq) + 2NaF(aq) → SrF₂(s) + 2NaCl(aq)

This means

1 mole of SrCl₂ is required to completely react with 2 moles of NaF

Now, we will determine the number of moles of SrCl₂ present

From the given information

Volume of SrCl₂ = 439 mL = 0.439 L

Concentration of SrCl₂ = 0.660 M

Using the formula

Number of moles = Concentration × Volume

∴Number of moles of SrCl₂ present = 0.660 × 0.439  

Number of moles of SrCl₂ present = 0.28974 mole

Since

1 mole of SrCl₂ is required to completely react with 2 moles of NaF

Then,

0.28974 mole of SrCl₂ will completely react with 2×0.28974 moles of NaF

2×0.28974 = 0.57948 mole

∴ The number of moles of NaF required to react completely with the SrCl₂ is 0.57948 mole

Now, for the volume of 0.510 M NaF required

Using the formula

$Volume = \frac{Number\ of\ moles}{Concentration}$

∴ Volume of NaF solution required = $\frac{0.57948}{0.510}$

Volume of NaF solution required = 1.136235 L

Volume of NaF solution required = 1136.235 mL

Volume of NaF solution required ≅ 1136 mL

Hence, the volume of the NaF solution required to react completely with the given SrCl₂ is 1136 mL

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