The balanced equation for the above reaction is
2Al + 6H₂O ---> 2Al(OH)₃ + 3H₂
stoichiometry of Al to H₂ is 2:3
number of Al moles reacted - 78.33 g / 27 g/mol = 2.901 mol
according to molar ratio
2 mol of Al forms - 3 mol of H₂
therefore 2.901 mol of Al - forms 3/2 x 2.901 = 4.352 mol
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP
if 1 mol occupies 22.4 L
then 4.352 mol occupies - 22.4 L/mol x 4.352 mol = 97.48 L
volume occupied by H₂ is 97.48 L
Answer:
a Control Variable in an experiment remains the same.
Answer:
a. <em><u>NH3</u></em>
<em><u>b</u></em><em><u>.</u></em>
<em><u>c.</u></em><em><u> </u></em><em><u>NaCl</u></em>
<em><u>d</u></em><em><u>.</u></em>
<em><u>e</u></em><em><u>.</u></em><em><u> </u></em><em><u>HCl</u></em>
<em><u>f</u></em><em><u>.</u></em><em><u> </u></em>