8. Based on the balanced equation below: 2 HNO3 1 Sr(OH)2 2 H2O + 1 Sr(NO3)2 + H

Question

8. Based on the balanced equation below: 2 HNO3 1 Sr(OH)2 2 H2O + 1 Sr(NO3)2 + How many mL of water would be formed from 2250mL of 10M strontium hydroxide reacting with excess nitric acid, if water has a density of 0.995g/mL? a. How many mL of water would be formed from 375mL of 10M nitric acid reacting with excess strontium hydroxide, if water has a density of 0.995g/mL? b. c. What would be the resultant nnolanty of strontium nitrate solution if OOrnL of 7M strontium hydroxide and 100mL of 7M nitric acid solutions were mixed (reacted)? (assuming negligible precipitate volume) d. What would be the resultant molarity of strontium nitrate solution if 300mL of 6M strontium hydroxide and 450mL of 9M nitric acid solutions were mixed (reacted)? (assuming negligible precipitate volume) Page 4

Explanation / Answer

a)

Volume of Sr(OH)2 = 2250 mL = 2.250 L, Concentration, Molarity = 10M

No of moles of Sr(OH)2 = Molarity X volume in litres

= 10 moles / L X 2.250 L

= 22.50 moles

According to equation for 1 mole of Sr(OH)2 2 moles of water formed. Thus for 22.50 moles of

Sr(OH)2 22.50X 2 = 45 moles of water formed.

No. of moles = Mass / molecular weight

Thus mass = Molecular weight of water X No. of moles of water = 18 g/mol X 45 mol

= 810 g

Volume of water = Mass / density = 810 g/ 0.995 g/ mL = 814.0 mL

b)

Volume of nitric acid = 375 mL,0.375 L Concentration, Molarity = 10M

No of moles of nitric acid = Molarity X volume in litres

= 10 moles / L X 0.375 L

= 3.75 moles

According to the equation for 2 moles of nitric acid gives 2 moles of water. Thus for 3.75 moles of

Nitric acid gives 3.75 moles of water.

No. of moles = Mass / molecular weight

Thus mass = Molecular weight of water X No. of moles of water = 18 g/mol X 3.75 mol

= 67.5 g

Volume of water = Mass / density = 67.5 g/ 0.995 g/ mL = 67.84 mL

c)

Molarity of nitric acid = 7 M, Volume = 100 mL, 0.100L

Molarity of Sr(OH)2 = 7 M, Volume = 100 mL, 0.100L

Total volume = 200 mL = 0.200 L

No of moles of nitric acid = Molarity X total volume in liters

= (7 mol/L) X 0.100 L = 0.7 mol

No of moles Sr(OH)2 = Molarity X total volume in litres

= (7 mol/L) X 0.100 L = 0.7 mol

1mole Sr(OH)2 reacts with 2 moles nitric acid to form 1 mole of Sr(NO3)2

0.35 mole Sr(OH)2 reacts with 0.7 moles nitric acid to form 0.35 mole of Sr(NO3)2

The molarity of Sr(NO3)2 in 0.750 L = no. of moles / 0.200 = 0.35moles / 0.200 L = 1.75 mol/ L = 1.75 M

d)

Molarity of nitric acid = 9 M, Volume = 450 mL = 0.450 L

Molarity of Sr(OH)2 = 6M, Volume = 300 mL = 0.300 L

Total volume in litres = 0.450 + 0.300 = 0.750 L

No of moles of nitric acid = Molarity X total volume in litres

= (9 mol/L) X 0.450 L = 4.05 mol

No of moles Sr(OH)2 = Molarity X total volume in litres

= (6 mol/L) X 0.300 L = 1.8 mol

1mole Sr(OH)2 reacts with 2 moles nitric acid to form 1 mole of Sr(NO3)2

1.8 mole Sr(OH)2 reacts with 3.6 moles nitric acid to form 1.8 mole of Sr(NO3)2

The molarity of Sr(NO3)2 in 0.750 L = no. of moles / 0.750 = 1.8 moles / 0.750 L = 2.4 mol/ L = 2.4 M

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