Question

3. Calcium carbonate reacts with HCl according to the following reaction: CaCO3 + 2 HCl  H2 + CO2 + CaCl2 How many grams of CaCO3 are required to react with 14.3 g HCl? Molar masses: HCl 36.5 g/mol; CaCO3 100.09 g/mol;

4.How many grams of solid NaOH must be weighed to make 300 mL of 0.145 M solution? NaOH

40.0g/mol

Answer 1

3) CaCO3 + 2 HCl --------------------> H2 + CO2 + CaCl2

1 mol 2 mol

100.09 g 2 x36.5 = 73 g

? 14.3 g

From the balanced equation, we can say that

100.09 g of CaCO3 is  required to react with 73 g HCl.

So, ? g  of CaCO3 is  required to react with 14.3 g HCl.

? = 100.09 g of CaCO3 x 14.3 g HCl /  73 g HCl

= 19.6 g of CaCO3

Therefore, 19.6 grams of CaCO3 are required to react with 14.3 g HCl.

4)

Molarity = (mass/ molar mass) x (1/ volume of solution in L)

Mass of solid NaOH = molarity x molar mass x volume of solution in L

= 0.145 M x 40.0 g/mol x 0.3 L

   = 1.74 g

Therefore,

1.74 grams of solid NaOH must be weighed to make 300 mL of 0.145 M solution.