How many grams of BaCl2 are formed when 35.00 mL of 0.00237 M Ba (OH) 2 reacts with excess Cl2 gas? 2 Ba (OH) 2 (aq) + 2 Cl2 (g) → Ba (OCl) 2 (aq) + BaCl2 (s) + 2 H2O (l)

0.0071g

Explanation:

From the question, we know that the molarity of the BaCl2 is 0.00237M. This means there are 0.00237 moles in 1dm^3 or 1000cm^3 of solution.

We also know that 35ml of the BaCl2 reacted. Here, we need to calculate the number of moles in 35.7ml of BaCl2.

This is calculated as follows;

0.00237moles are in 1000cm^3

Thus x moles will be present in 35ml (we should note that cm^3 is same as ml)

X = (0.00237 * 35) : 1000 = 0.00008295 moles.

From the reaction equation, we can see that 2 moles of BaCl2 yielded 1 mole of Ba (OH) 2.

This means 0.00008295mole of BaCl2 will yield 0.00008295 : 2 = 0.000041475 moles of Ba (OH) 2.

To calculate the mass of Ba (OH) 2 formed, we simple multiply the number of moles yielded by the molar mass of Ba (OH) 2.

Molar mass of Ba (OH) 2 = 137 + 2 (17)

= 171g/mol

Mass = 171 * 0.000041475 = 0.007092225g