Hello, Zoe,
The balanced equation for the Al reaction tells us we need 4 moles of Al for every 2 moles of Al2O3. Let's determine how many moles 3.4 grams of Al2O3 will make. The molar mass for Al2O3 is 102 g/mole.
That means we have 3.4g/(102g/mole), or 0.333 moles of the compound. (Ignoring sig figs until the end).
But the equation says we need twice as many moles of the Al (4 moles Al/2 moles Al2O3). So we need to start with 0.067 moles of Al. This translates into (0.067 moles) * (27g/mole Al) = 1.8 grams of Al.
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The equation Zn(s)+H20(g)-> ZnO(s)+H2(g) is balanced.
4.16 grams of Zn is 0.064 moles Zn (4.16g/(65.4 g/mole)). We get 1 mole of H2 for every mole of Al. Therefore we produce 0.64 moles of H2. That translates into 0.127 grams of H2 (0.0636 moles * 2g/mole).
Not much, but a lot of fun.
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Magnesium combustion with oxygen needs a balanced equation. The product is magnesium oxide, MgO
The initial equation is Mg + O2 = MgO
But we see that isn't balanced. If we add a coefficient of 2 to the Mg, it becomes balanced.
2Mg + O2 = 2MgO
We have to assume that oxygen is unlimited and that all the magnesium is consumed. The molar masses (in g/mole) are:
Mg: 24.3 g/mole
O2: 32
MgO: 40.3
For the 5.2 g and 0.84 g Mg samples, the moles are
5.2 grams: 5.2g/(24.3g/mole) = 0.214 moles
0.84g: 0.84g/(24.3 g.mole) = 0.035 moles
We get 2 moles of MgO for every 2 moles Mg. That's a 1:1 ratio.
Therefore we get 0.214 and 0.035 moles of MgO for the two reactions. Multiply the moles times the molar mass of MgO to get the two product masses:
0.214 moles * 40.3g/mole = 8.6g (2 sig figs) of MgO, and
0.035 moles * 40.3g/mole = 1.4g of MgO
I hope this helps,
Bob
