C, 0.746 mol Ag.
1 mol Ag = 6.022 x 1023 atoms of Ag -> 4.49 x 1023 atoms of Ag x 1mol Ag/6.022 x 1023 atoms -> 0.746 mol Ag
The mass of chlorine that react with 9.00 g of Al to form AlCl3 is 35.465 grams
Explanation
write the equation for reaction
that is
2 Al + 3 Cl2 = 2 Al CL3
find the moles of Al reacted
moles = mass/molar mass
9 g/ 27 g/mol = 0.333 moles of Al
by use of mole ratio between Al to Cl2 which is 2:3 find the moles of Cl2
mole of cl2 = 0.333 x3/2 = 0.4995 moles
mass of Cl2 is therefore = moles x molar mass
= 0.4995 x71 = 35.465 moles
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