Answer:
95.55% w/w Mg; 2.89% w/w Al
Explanation:
We can solve this question using both precipitates mixture. for example, for the oxide:
Mass MgO + Mass Al2O3 = 1.0022g
Moles MgO*40.3044g/mol + Moles Al2O3*101.96g/mol = 1.0022g
<em>Where we are writting the molar mass of each oxide. </em>
We can write, thus:
40.3044X + 50.98Y = 1.0022 <em>(1)</em>
<em>Where X are moles of Mg and Y moles of Al</em>
<em>Because 1 mole of Al2O3 are 2 moles of Al</em>
And for the other mixtue:
312.605X + 459.4317Y = 7.8154 <em>(2)</em>
<em></em>
Replacing (2) in (1):
312.605(1.0022-50.98Y / 40.3044) + 459.4317Y = 7.8154
312.605(0.024866-1.26487Y) + 459.4317Y = 7.8154
7.7732 - 395.406Y + 459.4317Y = 7.8154
64.0257Y = 0.0422
Y = 6.591x10⁻⁴ moles = Moles Al
The moles of Mg are:
312.605X + 459.4317*6.591x10⁻⁴ moles = 7.8154
312.605X + 0.3028 = 7.8154
312.605X = 7.5126
X = 0.02403 Moles = Moles Mg
The mass of each metal and its mass percent is:
Mg: 0.02403moles * (24.305g/mol) = 0.5841g / 0.6113g * 100 =
<h3>95.55% w/w Mg</h3><h3 />
Al: 6.591x10⁻⁴ moles * (26.98g/mol) = 0.01767g / 0.6113g * 100 =
<h3>2.89% w/w Al</h3>