Please answer ALL questions. Thank you. In the laboratory you dissolve 19.4 g of

Question

Please answer ALL questions. Thank you.

In the laboratory you dissolve 19.4 g of ammonium bromide in a volumetric flask and add water to a total volume of 125 mL What is the molarity o the solution? What is the concentration of the ammonium cation? What is the concentration of the bromide anion? M. M. In the laboratory you dissolve 23.3 g of ammonium sulfide in a volumetric flask and add water to a total volume of 250 mL What is the molarity of the solution? What is the concentration of the ammonium cation? What is the concentration of the sulfide anion? 3. For the following reaction, 20.2 grams of nitrogen monoxide are allowed to react with 4.50 grams of hydrogen gas. nitrogen monoxide (g) + hydrogen (g)-nitrogen (g) + water (I) What is the maximum amount of nitrogen gas 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 4 For the following reaction, 9.10 grams of propane (C3Hg) are allowed to react with 17.9 grams of oxygen gas. propane (C3Hg) (g)+ oxygen (g) carbon dioxide (g) +water (g) What is the maximum amount of carbon dioxide that can be formed? What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete?

Explanation / Answer

1)

a)

Molar mass of NH4Br,

MM = 1*MM(N) + 4*MM(H) + 1*MM(Br)

= 1*14.01 + 4*1.008 + 1*79.9

= 97.942 g/mol

mass(NH4Br)= 19.4 g

number of mol of NH4Br,

n = mass of NH4Br/molar mass of NH4Br

=(19.4 g)/(97.942 g/mol)

= 0.1981 mol

volume , V = 125 mL

= 0.125 L

Molarity,

M = number of mol / volume in L

= 0.1981/0.125

= 1.59 M

b)

[NH4+] = [NH4Br] = 1.59 M

Answer: 1.59 M

c)

[Br-] = [NH4Br] = 1.59 M

Answer: 1.59 M

Only 1 question at a time please. Questions are very lengthy

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