The answer to your question is below
Explanation:
Data
3.- grams of reactant
volume = 1l
a) 0.5 M NaCl
-Calculate the number of moles
moles = Molarity x volume
moles = 0.5 x 1
moles = 0.5
-Calculate the molar mass of NaCl
NaCl = 23 + 35.5 = 58.5 g
58.5g 1 mol
x 0.5 mol
x = (0.5 x 58.5) / 1
x = 29.25 grams of NaCl
b)
-Calculate the moles
moles = 2 x 1
moles = 2
-Calculate the molar mass of CaCl₂
CaCl₂ = 40 + 71 = 142 g
-Calculate the grams
142 g 1 mol
x 2 moles
x = (2 x 142) / 1
x = 284 grams of CaCl₂
c) 1.5 M NaOH
-Calculate the moles of NaOH
moles = 1.5 x 1
moles = 1.5
-Calculate the molar mass of NaOH
NaOH = 23 + 16 + 1 = 40 g
-Calculate the grams
40 g 1 mol
x 1.5 moles
x = (1.5 x 40)
x = 60 grams of NaoH
4.- mass of fructose = ?
Volume = 1l
Molarity = 0.75
-Calculate the moles of fructose
moles = 0.75 x 1
moles = 0.75
-Calculate the molar mass of fructose
C₆H₁₂O₆ = (12 x 6) + (12 x 1) + (16 x 6) = 72 + 12 + 96
= 180 g
-Calculate the grams of fructose
180 g 1 mol
x 0.75 moles
x = (0.75 x 180) / 1
x = 135 g of fructose
5.- Balanced chemical reaction
3Ba(NO₃)₂ + Al₂(SO₄)₃ ⇒ 3BaSO₄ + 2Al(NO₃)₃
-Calculate the moles of Al₂(SO₄)₃
moles = molarity x volume
moles = 0.35 x 0.025
= 0.00875
-Use proportions to calculate the moles of Ba(NO₃)₂
3 moles of Ba(NO₃)₂ 1 mol of Al(SO₄)₃
x 0.00875 moles of Al(SO₄)₃
x = (0.00875 x 3) / 1
x = 0.02625
-Calculate the volume of 0.280 M solution of Ba(NO₃)₂
volume = moles / molarity
volume = 0.02625 / 0.280
volume = 0.0938 l or 93.8 ml
5.- Balanced chemical reaction
2AgNO₃ + CaBr₂ ⇒ 2AgBr + Ca(NO₃)₂
- Calculate the moles of CaBr₂
moles = 0.128 x 0.035 = 0.0045
- Calculate the moles of AgNO₃
2 moles of AgNO₃ 1 mol of CaBr₂
x 0.0045 moles
x = (0.0045 x 2) /1
x = 0.009 moles
- Calculate the volume of AgNO₃
Volume = moles / Molarity
= 0.009 / 0.415
= 0.0215 l or 21.6 ml
