Question

Please Show working for the following clearly

4.) Barium hydroxide is reacted with calcium iodide. Calculate how many mL of .30M Ba (OH)2 needed to react with 250mL of .50M Cal2 (Write balanced equation first).

5.) Given the equation; N2 (g) + 3H2(g) (arrow) 2NH3(g). If 7.00L of N2 ( at STP) reacts with 15.0 L of H2, A.) How many L of NH3 is formed? B.) How many L of the excess gas remains unreacted?

C.) How many grams of NH3 are formed?

A.)

B.)

C.)

6.) In the titration, HBr + KOH (arrow) H2O + KBr, if 48.5 mL of 1.5M HBr reached the end point of 50.0mL of the KOH, calculate the molarity of the KOH.

Answer 1

Answer – 4) We are given, [Ba(OH)₂] = 0.30 M , [CaI₂] = 0.50 M ,

volume of CaI₂ = 250 mL , volume of Ba(OH)₂ = ?

First we need to write the balanced reaction –

Ba(OH)₂ + CaI₂ -----> BaI₂ + Ca(OH)₂

Now we need to calculate the moles of CaI₂

So, mole of CaI₂ = 0.50 M * 0.250 L

                           = 0.125 moles

From the balanced reaction

1 moles of CaI₂ = 1 mole of Ba(OH)₂

So, 0.125 moles of CaI₂ = 0.125 mole of Ba(OH)₂

Now we need to calculate the volume form the moles and molarity

We know, molarity = moles / volume (L)

So, volume (L) = moles / molarity

                        = 0.125 moles / 0.30 M

                       = 0.417 L

                        = 417 mL

So, 417 mL of .30M Ba (OH)2 needed to react with 250mL of .50M Cal2.

5) We are given, reaction

N₂(g) + 3H₂(g) ------> 2NH₃(g)

At , STP , volume of N₂(g) = 7.0 L and volume of H₂(g) = 15.0 L

A) First we need to calculate moles of each reactant

we know, at STP 22.4 L = 1 mol

so, moles of N₂(g) = 7.0 L *1 mole / 22.4 L = 0.312 moles

moles of H₂(g) = 15.0 L *1 mole / 22.4 L = 0.669 moles

from the balanced reaction - moles of NH₃ from N₂

1 mole of N₂(g) = 2 moles of NH₃(g)

So, 0.312 moles of N₂(g) = ?

= 0.625 moles of NH₃

From H₂

3 mole of H₂(g) = 2 moles of NH₃(g)

So, 0.669 moles of H₂(g) = ?

= 0.446 moles of NH₃

So limiting reactant is H₂ and moles of NH₃ = 0.446 moles

So, volume of NH₃ = 0.446 moles * 22.4 L

                                = 10.0 L

B) Moles of excess gas

3 mole of H₂(g) = moles of N₂(g)

So, 0.669 moles of H₂(g) = ?

= 0.223 moles of NH₃

So, excess moles of gas, N₂ = 0.312 – 0.223

                                             = 0.0892 moles

So, volume of excess gas remains unreacted, N₂ = 0.0892 moles * 22.4 L / 1mole

                                                                              = 2.0 L

C) Moles of NH₃ formed = 0.446 moles

so, mass of NH₃ = 0.446 moles * 17.0307 g/mol

                           = 7.60 g of NH₃

6) We are given,

HBr + KOH ----> H2O + KBr

[HBr] = 1.5 M , volume = 48.5 mL , volume of KOH = 50.0 mL

Now we need to calculate the moles of HBr

Moles of HBr = 1.5 M * 0.0485 L

                    = 0.0727 moles

So, from the reaction

1 moles of HBr = 1 moles of KOH

So, moles of KOH = 0.0727 moles

So, molarity of KOH = 0.0727 moles / 0.050 L

                               = 1.46 M