Question

1) According to the following reaction, how many grams of hydrogen gas are necessary to form 0.603 moles ammonia? nitrogen (g) + hydrogen (g) ammonia (g)

2) For the following reaction, 6.52 grams of methane (CH4) are allowed to react with 29.8 grams of carbon tetrachloride. methane (CH4) (g) + carbon tetrachloride (g) dichloromethane (CH2Cl2) (g) What is the maximum amount of dichloromethane (CH2Cl2) that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

3) You need to make an aqueous solution of 0.130 M nickel(II) bromide for an experiment in lab, using a 500 mL volumetric flask. How much solid nickel(II) bromide should you add?

4)How many milliliters of an aqueous solution of 0.155 M barium acetate is needed to obtain 9.17 grams of the salt?

5) In the laboratory you dissolve 13.0 g of nickel(II) bromide in a volumetric flask and add water to a total volume of 500 mL. What is the molarity of the solution? M. What is the concentration of the nickel cation? M. What is the concentration of the bromide anion? M

6) In the laboratory you dissolve 15.5 g of barium acetate in a volumetric flask and add water to a total volume of 250 mL. What is the molarity of the solution? M. What is the concentration of the barium cation? M. What is the concentration of the acetate anion? M.

Answer 1

1) Consider a reaction N_{2 (g)} + 3 H_2(g) 2 NH_3(g)

From reaction , 3 mole hydrogen react to produce 2 mole ammonia. Hence , to produce 0.603 mole ammonia moles of hydrogen required is ( 0.603 x 3 / 2 = 0.9045 mole ammonia) .

ANSWER: 0.9045 mole hydrogen required.

2) Consider a reaction CH_4(g) + CCl_4(g) 2 CH₂Cl_2(g)

First calculate moles of methane and carbon tetrachloride

Molar mass of CH₄ = 12+ ( 4 x 1) = 16 g / mol

Molar mass of CCl₄ = 12 + ( 4 x 35.45 ) = 153.8 g / mol

Moles of CH₄ = mass / molar mass

= 6.52 g / (16 g / mol)

= 0.4075 mol

Moles of CCl₄ = mass / molar mass

= 29.8 g / (153.8 g / mol)

= 0.1938  mol

Provided ratio of CH₄ : CCl₄ = 0.4075 mol : 0.1938  mol = 2.1 : 1

From reaction , ratio of CH₄ : CCl₄ = 1 mol : 1 mol = 1 : 1

Comparing above two ratios , CCl₄ is present in less amount and is a limiting reactant.

Therefore, amount of CH₂Cl₂ will depend upon amount of  CH_4.

From reaction 1 mole CCl₄ 2 mole CH₂Cl₂

153.8  g CCl₄ 2 x 84.9 g CH₂Cl₂

Hence, 29.8 g CCl₄ ( 2 x 84.9 x 29.8 / 153.8 ) g CH₂Cl₂

32.9 g CH₂Cl₂

Maximum amount of CH₂Cl₂ = 32.9 g

Amount of excess reagent :

Excess moles of methane remained unreacted = moles of methane - moles of carbon tetrachloride

= 0.4075 - 0.1938

= 0.2137 mole

0.2137 mole methane = 0.2137 mol x ( 16 g / 1 mol ) = 3.419 g

Excess methane = 3.419 g

3 ) We know that , Molarity = no of moles / volume of solution in L

Therefore, no of moles of Nickel(II) bromide = molarity x volume of solution in L

= 0.130 mol / L x 0.5 L

= 0.065 mol

also, No of moles = mass / molar mass

Molar mass of Nickel(II) bromide = 58.69 + ( 2 x 79.90) = 218.5 g / mol

Therefore,mass of Nickel(II) bromide =   No of moles x molar mass

= 0.065 mol x 218.5 g / mol  

=14.20 g

ANSWER: mass of Nickel(II) bromide = 14.20 g

4) We know that , Molarity = no of moles / volume of solution in L

No of moles = mass / molar mass

molar mass of Ba( CH3COO)2= 137.33 + ( 4 x 12 ) + ( 6 x 1) +( 4 x 16) = 255.33 g / mol

Therefore, no of moles = 9.17 g / 255.33 g / mol

= 0.0359 mol

From molarity formula we can write,

volume of solution in L = no of moles / molarity

= 0.0359 mol / 0.155 mol / L

= 0.232 L

= 232 ml

ANSWER: volume of solution = 232 ml

5 ) Molar mass of Nickel(II) bromide = 58.69 + ( 2 x 79.90) = 218.5 g / mol

No of moles = mass / molar mass

Hence, no of moles of Nickel(II) bromide = 13.0 g / 218.5 g / mol

= 0.0595 mol

We know that, molarity = no of moles / volume of solution in L

Hence, molarity Nickel(II) bromide = 0.0595 mol / 0.5 L

= 0.119 mol / L

Nickel(II) bromide contain 1 Ni (II) and 2 bromide ions per molecule.

Hence [ Ni(II) ] = [ Nickel(II) bromide ] = 0.119 mol / L  

[ Br - ] = 2 x [ Nickel(II) bromide ] = 2 x 0.119 mol / L = 0.238 mol / L

6) molar mass of Ba( CH3COO)2= 137.33 + ( 4 x 12 ) + ( 6 x 1) +( 4 x 16) = 255.33 g / mol

No of moles = mass / molar mass

Hence, no of moles of barium acetate = 15.5 g / 255.33 g / mol

= 0.0607 mol

We know that , Molarity = no of moles / volume of solution in L

= 0.0607 mol / 0.250 L

= 0.243 mol / L

barium acetate contain 1 Ba (II) and 2 acetate ions per molecule.

[ barium acetate ] = [Ba (II)]= 0.243 mol / L

[ acetate] = 2 x [ barium acetate ] = 2 x 0.243 mol / L =0.486 mol / L