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vanessa791
13.02.2021 •
Chemistry
How many grams of Al2O3 are needed to produce 24.5 L of O2 gas at STP?
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Ответ:
74.35 g
Explanation:
The equation of the reaction is given as;
2Al2O3 --> 4Al + 3O2
From the reaction;
2 mol of Al2O3 reacts to produce 3 mol of O2
Converting 2 mol of Al2O3 to mass;
Number of moles = Mass / Molar mass
Mass = Number of moles * Molar mass
Mass = 2 * 101.96 = 203.92 g
Converting 3 mol of O2 to volume;
1 mol = 22.4 L
3 mol = x
x = 22.4 * 3 = 67.2 L
This means 203.92g of Al2O3 produces 67.2L of O2
xg would produce 24.5 L of O2
203.92 = 67.2
x = 24.5
x = (24.5 * 203.92) / 67.2
x = 74.35 g
Ответ:
they use symbols to represent land features.
Explanation: