Odpowiedź:
48g 80g
[tex]2Mg+O_2-- > 2MgO[/tex]
12g xg
[tex]M_M_g=24\frac{g}{mol} \\M_M_g_O=40\frac{g}{mol}[/tex]
48gMg----- 80gMgO
12gMg----- xg MgO
48x=80*12
48x=960 /:48
x=20g MgO
n=[tex]\frac{20\frac{g}{mol} }{40\frac{g}{mol} } =\frac{1}{2}mola[/tex]MgO
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Odpowiedź:
48g 80g
[tex]2Mg+O_2-- > 2MgO[/tex]
12g xg
[tex]M_M_g=24\frac{g}{mol} \\M_M_g_O=40\frac{g}{mol}[/tex]
48gMg----- 80gMgO
12gMg----- xg MgO
48x=80*12
48x=960 /:48
x=20g MgO
n=[tex]\frac{20\frac{g}{mol} }{40\frac{g}{mol} } =\frac{1}{2}mola[/tex]MgO