Calculate the masses of ca(no3)2•4h2o(s) and kio3(s) required to make 10.0 g of ca(io3)2(s)

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

Q2eqefefgvsdfvfefsvsdvcwef