6. Silver metal reacts with concentrated nitric acid (HNO3) to produce the red-b

Question

6. Silver metal reacts with concentrated nitric acid (HNO3) to produce the red-brown gas nitrogen dioxide, silver nitrate, and water. Write the balanced equation for the reaction. How many grams of nitric acid are required to use up 5.00 g Ag? How many grams of nitrogen dioxide would be formed a. c. 7. The combustion of iron (II) sulfide produces iron (II) oxide and sulfur dioxide. Write the balanced equation for the reaction. Starting with 0.48 mole of iron (II) sulfide and excess oxygen, how many grams of iron (III) oxide can be formed? 8. Nitrogen dioxide can react with water to form nitric acid and nitrogen monoxide. Write the balanced equation for the reaction. How many moles of water are required to use up 230, g of nitrogen monoxide? a. c.How many milligrams of nitrogen monoxide can be formed from 2.4 x 1021 molecules of nitrogen dioxide?

Explanation / Answer

6.

a)    The balanced equation is

Ag + 2 HNO3 --------------------- AgNO3 + NO2 + H2O

b) mass of Ag = 5.00 grams

molar mass of HNO3 = 63 gram/mole

molar mass of Ag = 108 grams

According to equation

1 mole of Ag = 2 mole of HNO3

108 grams of Ag = 2x63 grams of HNO3

5.00 grams of Ag = ?

                               = 2 x63x5.00/108 =5.83 gras

So amount of HNO3 requird = 5.83 grams.

7.

a) 4 FeS + 7 O2 ---------------- 2 Fe2O3 + 4 SO2

b)number of moles of FeS= 0.48 moles

According to equation

4 moles of FeS = 2 moles of Fe2O3

0.48 moles of FeS = ?

                               = 2 x 0.48/4 = 0.24 moles

number of moles o f Fe2O3 formed = 0.24 moles

molar mass of Fe2O3 = 159.7 gram/mole

mass of 0.24 moles of Fe2O3 = 0.24 x 159.7 = 38.328 grams

mass of Fe2O3 formed = 38.33 grams.

8)

a) 3 NO2   + H2O --------------- 2 HNO3 + NO

b)

According to equation

1 mole of H2O = 1 mole of NO

molar mass of H2O = 18 grams/mole

molar mass of NO = 30 grams/mole

30 grams of NO = 18 grams of H2O

230.0 grams of NO = ?

                              = 18 x 230 /30 = 138 grams of H2O

mass of H2O required = 138 grams

number of moles of H2O = 138/18 = 7.67 moles

number of moles of H2O = 7.67moles.

c)According to equation

3 moles of NO2 = 1 moles of NO

3 x 6.023x10^23 molecules of NO2= 30 grams of NO

2.4x10^21 molecules of NO2 = ?

                                             = 30 x 2.4 x10^21/3x6.023x10^23 = 3.98 x 10-2

mass of NO formed = 3.98 x 10^-2 grams = 3.98 x 10^-2 x 10^3 mg = 39.8 mg.

mass of NO formed = 39.8 mg

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