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