Record the amount of solute left after one minute of stirring for each solution. Beaker Amount of solute at start (g)(first number beside A, B, or C) Amount of solute at end (g)(second number beside A, B or C) Solution with Compound A 50 0 Solution with Compound B 50 15 Plain water in Beaker C 0 0 Variables – Procedure 1 1. What was the independent variable in Procedure 1? (1 point) Solution A, B, and C 2. What was the dependent variable in Procedure 1? (1 point) The boiling point 3. List three (3) controlled variables in Procedure 1. (1 point) (Hint: NOT Beaker C with the plain water)

he would use the colour black

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

you gotta add the following molecules so we can give u an answer

explanation: