"Mg + 2HCl \\to MgCl_2 + H_2"
1 mole of Mg reacts with 2 moles of HCl
24g of Mg = 2(36.5)g of HCl
24g of Mg = 73g of HCl
4g of Mg = x of HCl
x =4×73/24
x = 12.17g
Therefore, 12.17g of Mg is produced =
12.17g/24g/mol = 0.507 mol
To get the volume needed,
Volume = no. of moles/concentration
Volume = 0.507/1.85 = 0.274dm³ = 0.274L.
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