Question

How many moles of Na2HPO4 are present in 469.6 g of Na2HPO4? Question 6 How many moles of caffeine, C8H10N402, are present in 191.11 g of caffeine? Question 7 How many grams of FeCl3 can be prepared from 5.01 grams of Fe and excess Cl2 when the following reaction is balanced? _Fe (s) + _ Cl2@ FeCl3 (1) (unbalanced) Question 8 Calculate the molarity of a solution prepared by mixing 38.5 g of CH3CH2OH with enough water to give 500 mL of solution

Answer 1

5. molar mass of Na₂HPO₄= 141.96 g/mol

=>moles of Na₂HPO₄= gram wt/molar mass

=>moles of Na₂HPO₄=469.6g/ 141.96 g/mol

=>moles of Na₂HPO₄= 3.31 mol

6. molar mass of caffeine= 194.19 g/mol

=>moles of caffeine= gram wt/molar mass

=>moles of caffeine=191.11g/ 194.19 g/mol

=>moles of caffeine= 0.98 mol

7. Balance chemical equation

2Fe+3Cl₂===>2FeCl₃

moles of Fe= 5.01g/ 55.845 g/mol

moles of Fe= 0.090 mol

from balanced equation

1 mole of Fe yields= 1 mol of FeCl₃

=> moles of FeCl₃= 0.090 mol

molar mass of FeCl₃= 162.2 g/mol

=> amount of FeCl₃ in gram= mole× molar mass

=> amount of FeCl₃ in gram= 0.090 mol×162.2g/mol

=> amount of FeCl₃ in gram= 14.6 g

8. molarity= moles/volume (in ltr)

moles of ethanol= 38.5 g/ 46.07 g/mol

moles of ethanol=0.84 mol

volume= 500mL= 0.500 L

molarity = 0.84/0.500

=> molarity= 1.68M

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