Question

How many mL of 1.69 M HClO₃ will be needed to neutralize 177 mL of 4.61 M Sr(OH)₂.

When aqueous solutions of (NH₄)₂CrO₄ and Ba(NO₃ )₂ are combined, BaCrO₄ precipitates. Calculate the mass, in grams, of the BaCrO₄ produced when 1.52 mL of 0.116 M Ba(NO₃ )₂ and 4.56 mL of 0.622 M (NH₄)₂CrO₄ are mixed. Calculate the mass to 3 significant figures.

Answer 1

Answer – We are given, [HClO₃] = 1.69 M , [Sr(OH)₂] = 4.61 M

volume = 177 mL

We need to calculated volume of 1.69 M HClO₃

First, we need to write the reaction –

Reaction - 2 HClO₃ + Sr(OH)₂ ----> Sr(ClO₃)₂ + 2 H₂O

Now for calculating volume of HClO₃ we need moles of HClO₃ and moles of HClO₃ is calculated from the moles of Sr(OH)₂ using stoichiometry.

Moles of Sr(OH)₂ = molarity * volume (L)

                             = 4.61 M * 0.177 L

                            = 0.816 moles

From the balanced reaction –

1 moles of Sr(OH)₂ = 2 moles of HClO₃

so, 0.816 moles of Sr(OH)₂ = ? moles of HClO₃

= 1.63 moles of HClO₃

Now volume of HClO₃,

we know formula

Volume of HClO₃ = moles / molarity

                            = 1.63 moles / 1.69 M

                             = 0.965 L

                             = 965 mL

so, 965 mL of 1.69 M HClO₃ will be needed to neutralize 177 mL of 4.61 M Sr(OH)₂.

Part 2- We are given, [(NH₄)₂CrO₄] = 0.622 M volume of (NH₄)₂CrO₄ = 4.56 mL

[Ba(NO₃)₂] = 0.116 M , volume = 1.52 mL

Reaction –

(NH₄)₂CrO₄ (aq) + Ba(NO₃)₂ (aq) ----> BaCrO₄ (s) + 2 NH₄NO₃(aq)

Moles of each reactant –

Moles of (NH₄)₂CrO₄ (aq) = 0.662 M * 0.00456 L

                                                = 0.00284 moles

Moles of Ba(NO₃)₂ (aq) = 0.116 M * 0.00152 L

                                           = 0.000176 moles

Limiting reactant –

Moles of BaCrO₄ (s) from (NH₄)₂CrO₄ (aq)

From the balanced reaction –

1 moles of (NH₄)₂CrO₄ (aq) = 1 moles of BaCrO₄ (s)

so, 0.00284 moles of (NH₄)₂CrO₄ (aq) = ? moles of BaCrO₄ (s)

= 0.00284 moles of BaCrO₄ (s)

Moles of BaCrO₄ (s) from Ba(NO₃)₂ (aq)

From the balanced reaction –

1 moles of (Ba(NO₃)₂ (aq) = 1 moles of BaCrO₄ (s)

so, 0.00176 moles of Ba(NO₃)₂ (aq) = ? moles of BaCrO₄ (s)

= 0.00176 moles of BaCrO₄ (s)

The moles of BaCrO₄ (s) is lowest form the reactant (Ba(NO₃)₂ (aq) , so limiting reactant is (Ba(NO₃)₂ (aq) .

Moles of BaCrO₄ (s) = 0.000176 moles

Mass of BaCrO₄ (s) = 0.000176 moles x 253.37 g/mol

                                   = 11.3 g

so, 11.3 gram of the BaCrO₄ produced when 1.52 mL of 0.116 M Ba(NO₃)₂ and 4.56 mL of 0.622 M (NH₄)₂CrO₄ are mixed.