Question

. Calculate the masses of Ca(NO3)2•4H2O(s) and KIO3(s) required to make 10.0 g of Ca(IO3)2(s)

Answer 1

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

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

`\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}`

`\text{Number of moles}=(10.0g)/(390 g/mol)=0.025moles`

`Ca(NO_3)_2.4H_2O(s)+2KIO_3(s)\rightarrow Ca(IO_3)_2+2KNO_3+4H_2O`

1 mole of
`Ca(IO_3)_2` is formed from 1 mole of
`Ca(NO_3)_2.4H_2O` and 2 moles of
`KIO_3`

Thus 0.025 moles of
`Ca(IO_3)_2` is formed from 0.025 moles of
`Ca(NO_3)_2.4H_2O` and 0.05 moles of
`KIO_3`

Mass of
`Ca(IO_3)_2.4H_2O=moles* {\text {Molar mass}}=0.025* 236=5.9 grams`

Mass of
`KIO_3=moles* {\text {Molar mass}}=0.05* 214=10.7grams`

Answer 2

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


Step-by-step 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