Calculate the mass of NaCl required to prepare 360.0 mL of a 0.35 M solution.

Explanation:

Molar mass of NaCl = 58.44

Molar mass of NaCl(58.44g)  addding to 1000ml give rise to 1M solution always, this can be writtne as below

58.44 >1000 ml ---> 1M

X gram 360 ml ---0.35 M

X Gram= 58.44*360*0.35/1000*1 = 7.3 gram

hope you understood the solution

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