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

the term hydride is commonly named after binary compounds that hydrogen forms with other elements of the periodic table. hydride compounds in general form with almost any element, except a few noble gases. the trends and properties vary according to the type of intermolecular force that bonds the elements together, the temperature, its molecular masses, and other components. hydrides are classified into three major groups, depending on what elements the hydrogen bonds to. the three major groups are covalent, ionic, and metallic hydrides. formally, hydride is known as the negative ion of a hydrogen, h-, also called a hydride ion. because of this negative charge, hydrides have reducing, or basic properties. its special characteristics will be further discussed.

explanation: