Alab technician needs to create 570.0 milliliters of a 2.00 m solution of magnesium chloride (mgcl2). to make this solution, how many grams of magnesium chloride does the technician need? refer to the periodic table for . express your answer to three significant figures.

108.3g

Explanation:

Given parameters:

Volume of solution = 570mm = 0.57L (1000mm = 1L)

Molarity of solution = 2M

Unknown

Mass of MgCl₂

Solution

We first find the  number of moles in the given concentration of magnessium chloride using the expression below:

       Number of moles of MgCl₂= Molarity x volume = 2 x 0.57 = 1.14moles

Using this known moles, we can the unknown mass of MgCl₂ the technician would require:

       Mass of MgCl₂ required = number of moles of MgCl₂ x molar mass

Molar mass of MgCl₂:

Atomic mass of Cl = 35.5g

Atomic mass of Mg = 24g

MgCl₂ = 24 + (35.5x2) = 95gmol⁻¹

Mass of MgCl₂ required = 1.14mole x 95gmole⁻¹ = 108.3g

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2