Question

How many grams of CaCl2 are formed when 20.00 mL of 0.00237 M Ca(OH)2 reacts with excess Cl2 gas? 2 Ca(OH)2(aq) + 2 Cl2(g) → Ca(OCl)2(aq) + CaCl2(s) + 2 H2O(l)

Answer 1

The mass of CaCl2 formed when 20.00 mL of 0.00237 M Ca(OH)2 reacts with excess Cl2 gas is 0.00263 grams.

Step-by-step explanation:

To find the number of grams of CaCl2 formed, we need to calculate the moles of CaCl2 formed first. We can use the balanced chemical equation to determine the mole ratio between Ca(OH)2 and CaCl2.

According to the balanced equation, 2 moles of Ca(OH)2 reacts with 2 moles of Cl2 to form 1 mole of CaCl2. That means, for every mole of Ca(OH)2 consumed, we will get 0.5 moles of CaCl2.

Given that the concentration of Ca(OH)2 is 0.00237 M, which means there are 0.00237 moles of Ca(OH)2 in 1 liter or 1000 mL of solution.

So, to find the moles of Ca(OH)2 consumed, we can use the following equation:

moles of Ca(OH)2 consumed = concentration of Ca(OH)2 x volume of Ca(OH)2 solution (in liters)

moles of Ca(OH)2 consumed = 0.00237 M x (20.00 mL / 1000 mL) = 4.74 x 10-5 moles

Finally, we can find the moles of CaCl2 formed:

moles of CaCl2 formed = 0.5 x moles of Ca(OH)2 consumed = 0.5 x 4.74 x 10-5 moles = 2.37 x 10-5 moles

Since the molar mass of CaCl2 is 110.98 g/mol, we can calculate the mass of CaCl2 formed:

mass of CaCl2 formed = moles of CaCl2 formed x molar mass of CaCl2

mass of CaCl2 formed = 2.37 x 10-5 moles x 110.98 g/mol = 0.00263 g

Answer 2

0.026 grams of
`CaCl_2` will be produced.

From the balanced equation:

`2 Ca(OH)_2(aq) + 2 Cl_2(g) -- > Ca(OCl)_2(aq) + CaCl_2(s) + 2 H_2O(l)`

The mole ratio of
`Ca(OH)_2` to
`CaCl_2` is 2:1.

Mole of
`Ca(OH)_2` available = molarity x volume

= 0.00237 x 20/100 = 0.000474 moles

Equivalent mole of
`CaCl_2` that will be produced = 0.000474/2

= 0.000237 moles

Mass of 0.000237 moles
`CaCl_2` = mole x molar mass

= 0.000237 x 110.98

= 0.026 grams

In other words, 0.026 grams of
`CaCl_2` will be produced.