how many grams of mgcl2 will be produced from 16g of mg(oh)2 and 11g of hcl

Answer 1

We must first find the chemical equation for the reaction that applies.

2 HCl+Mg(OH)2---->MgCl2+ 2 H2O

Now, we must convert the reactants into moles. We divide the grams by molar mass for each.

Mg(OH)2: 16 g/(58.32 g/mol)=0.27 moles

HCl: 11 g/(36.46 g/mol)=0.3 moles

Now, we need to recognize that there is not a perfect ratio between these two reactants in terms of moles. So, there will be some excess reactant available at the end of the reaction.

The limiting reactant is HCl because it will be used up by the end of the experiment. Because the ratio of moles in this reaction in terms of HCl to Mg(OH)2 is 2 to 1. 0.3 moles of HCl, all of it, will react with 0.15 moles of Mg(OH)2.

Plus, for every two moles of HCl reacting, a mole of MgCl2 is a product. So, since 0.3 moles of HCl are reacting, 0.15 moles of MgCl2 is produced.

Now that we know this, we can find how many grams of MgCl2 will be produced. Let's multiply 0.15 with the molar mass of MgCl2.

0.15*95.21=14.28 g