Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
2al(s)+3cl2(g)→2alcl3(s)
you are given 10.0 g of aluminum and 15.0 g of chlorine gas. if you had excess chlorine, how many moles of of aluminum chloride could be produced from 10.0 g of aluminum?

0.3706 moles of aluminum chloride would be produced from 10.0 g of aluminum

Explanation:

chemical equation:

2Al(s) + 3Cl2(g) → 2AlCl3(s)

Given data:

mass of aluminium = 10 g

mass of chlorine = 15 g

First of all we will calculate the moles of each reactant:

moles of aluminium= 10 g/ 26.98 g/mol

moles of aluminium= o.3706 mol

moles of Cl2 = 15 g/ 10.90 g/mol

moles of Cl2 = 0.2116 mol

now we compare the moles aluminium with aluminium chloride

    Al     :  AlCl3

    2     :    2

    1      :     1

0.3706  :    0.3706

now we compare the moles chlorine with aluminium chloride

    Cl    :   AlCl3

     3    :      2

 0.2116  :   2/3× 0.2116 = 0.1412 mole

although chlorine is limiting reactant but in question we are given excess chlorine so 10 g of aluminium will produce 0.3706 moles of aluminium chloride.

mass of AlCl3 = 0.3706 mol × 133.34 g/mol

mass of AlCl3 = 50.13 g

the answer is b & c i believe : )

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer