Respuesta :

Edufirst
Answer: 6.1 g of Ca(NO₃)₂•4H₂O and 5.5 g of KIO₃


Explanation:


1) Calculate the number of moles of Ca(NO₃)₂ in 10.0 g


i) molar mass of Ca(NO₃)₂ = 40.1 g/mol + 2x126.9 g/mol + 2x3x16.0 g/mol = 389.9 g/mol


ii) Formula: number of moles = mass in grams / molar mass


number of moles = 10.0 g / 389.9 g/mol = 0.02565 moles of Ca(NO₃)₂


2) Write the chemical equation to state the mole ratio:


i) Ca(NO₃)2•4H₂O(s) + 2KIO₃(s) --> Ca(IO₃)₂(s) + 2KNO₃ + 4H₂O


ii) mole ratio: 1 mol Ca(NO₃)2•4H₂O(s) : 2 mol KIO₃(s) : 1 mol Ca(IO₃)₂(s)


3) Use proportionality to find the actual number of moles


i) Ca(NO₃)₂•4H₂O(s)

1 mol Ca(NO₃)₂•4H₂O(s) / 1 mol Ca(IO₃)₂ = x / 0.02565 mol Ca(IO₃)₂ => x = 0.02565 mol Ca(NO₃)₂•4H₂O(s)


molar mass of Ca(NO₃)₂•4H₂O(s) = 40.0g/mol + 2x14.0g/mol + 2x3x16.0g/mol + 4x18.0g/mol = 236.0g/mol


mass in grams = number of moles x molar mass = 0.02565 mol x 236.0 g/mol = 6.1 g


ii) KIO₃


1mol KIO₃/1molCa(IO₃)₂ = x / 0.02565 mol Ca(IO₃)₂ => x = 0.02565 mol KIO₃


molar mas of KIO₃ = 39.1 g/mol + 126.9 g/mol + 3x16.0 g/mol = 214.0 g/mol


mass in grams = 0.02565 mol x 214.0 g/mol = 5.5 g

Answer:  The masses of [tex]Ca(NO_3)_2.4H_2O[/tex] and [tex]KIO_3[/tex] required to make 10.0 g of [tex]Ca(IO_3)_2[/tex] is 5.9 and 10.7 grams respectively.

Explanation: To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}[/tex]

[tex]\text{Number of moles}=\frac{10.0g}{390 g/mol}=0.025moles[/tex]

[tex]Ca(NO_3)_2.4H_2O(s)+2KIO_3(s)\rightarrow Ca(IO_3)_2+2KNO_3+4H_2O[/tex]

1 mole of [tex]Ca(IO_3)_2[/tex] is formed from 1 mole of [tex]Ca(NO_3)_2.4H_2O[/tex] and 2 moles of [tex]KIO_3[/tex]

Thus 0.025 moles of [tex]Ca(IO_3)_2[/tex] is formed from 0.025 moles of [tex]Ca(NO_3)_2.4H_2O[/tex] and 0.05 moles of [tex]KIO_3[/tex]

Mass of [tex]Ca(IO_3)_2.4H_2O=moles\times {\text {Molar mass}}=0.025\times 236=5.9 grams[/tex]

Mass of [tex]KIO_3=moles\times {\text {Molar mass}}=0.05\times 214=10.7grams[/tex]

Otras preguntas