Question

2KHCO3(s)  K2CO3(s) + CO2(g) + H2O(l) How many moles of potassium carbonate will be produced if 454 g of potassium hydrogen carbonate are heated ? A) 2.27 mol B) 3.29 mol C) 11.4 mol D) 227 mol E) 4.54 mol

Answer 1

Answer:

A

Explanation:

gram mol

2KHCO3 K2CO3

200 1

454 X

X=454÷200=2.27g

Answer 2

Answer:

2.27 moles potassium carbonate will be produced. ( Option A is correct.)

Explanation:

Step 1: Data given

Mass of potassium hydrogen carbonate (KHCO3) = 454 grams

Molar mass KHCO3= 100.1 g/mol

Step 2: The balanced equation

2KHCO3(s) → K2CO3(s) + CO2(g) + H2O(l)

Step 3: Calculate moles KHCO3

Moles KHCO3 = mass KHCO3 / molar mass KHCO3

Moles KHCO3 = 454 grams / 100.1 g/mol

Moles KHCO3 = 4.54 moles

Step 4: Calculate moles of K2CO3

For 2 moles KHCO3 we'll have 1 mol K2CO3, 1 mol CO2 and 1 mol H2O

For 4.54 moles KHCO3 we'll have 4.54/2 = 2.27 moles K2CO3

2.27 moles potassium carbonate will be produced. ( Option A is correct.)