For this you need the Atomic Mass of Al. Take the number of grams and divide it by the atomic mass. Multiply it by one mole for units to cancel.
94.5 grams Al / (27.0 grams) = 3.50 moles Al
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
Fe2O3 + 2Al ===> Al2O3 + 2FeIn this reaction the number of moles of Al2O3 produced is dependent on the number of moles of Fe2O3 and Al that one starts with. For every 1 mole Fe2O3 and 2 moles Al, one gets 1 moles of Al2O3.
The balanced chemical equation for the reaction between aluminum (Al) and oxygen (O₂) to form aluminum oxide (Al₂O₃) is: [ 4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3 ] According to the balanced equation, 4 moles of aluminum (Al) produce 2 moles of aluminum oxide (Al₂O₃). Therefore, if 4.0 moles of aluminum completely react, it will produce ( \frac{2}{4} \times 4.0 ) moles of aluminum oxide. Calculate that to find the answer.
Sounds like an acid metal reaction to produce hydrogen gas, so I will assume you meant hydrochloric acid. Complete reaction indicates aluminum is limiting and drives the reaction. 2Al + 6HCl --> 2AlCl3 + 3H2 3.0 moles Al (3 moles H2/2 moles Al) = 4.5 mole hydrogen gas produced =========================
Since the ratio of moles of Al to moles of Al2O3 is 4:2, if 5.23 mol Al completely reacts, 2.615 mol Al2O3 can be made.
1mol Al = 26.981538g Al1mol Al atoms = 6.022 x 1023 atomsConvert grams to moles.1g Al x 1mol/26.981538g = 0.04mol AlConvert moles to atoms.0.04mol Al x 6.022 x 1023 atoms/mol = 2 x 1022 atoms Al
This reaction is:2 Al + 2 H2O + 2 NaOH = 2 NaAlO2 + 3 H2From 4 moles of Al 6 moles of hydrogen are obtained.
Al+HCl===> AlCl3+H2 Is the reaction. You need &.2 moles of HCl.
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
Fe2O3 + 2Al ===> Al2O3 + 2FeIn this reaction the number of moles of Al2O3 produced is dependent on the number of moles of Fe2O3 and Al that one starts with. For every 1 mole Fe2O3 and 2 moles Al, one gets 1 moles of Al2O3.
The answer is 3,375 moles oxygen.
9 grams aluminum (1 mole Al/26.98 grams) = 0.3 moles aluminum ==============
Sounds like an acid metal reaction to produce hydrogen gas, so I will assume you meant hydrochloric acid. Complete reaction indicates aluminum is limiting and drives the reaction. 2Al + 6HCl --> 2AlCl3 + 3H2 3.0 moles Al (3 moles H2/2 moles Al) = 4.5 mole hydrogen gas produced =========================
The balanced chemical equation for the reaction between aluminum (Al) and oxygen (O₂) to form aluminum oxide (Al₂O₃) is: [ 4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3 ] According to the balanced equation, 4 moles of aluminum (Al) produce 2 moles of aluminum oxide (Al₂O₃). Therefore, if 4.0 moles of aluminum completely react, it will produce ( \frac{2}{4} \times 4.0 ) moles of aluminum oxide. Calculate that to find the answer.
Since the ratio of moles of Al to moles of Al2O3 is 4:2, if 5.23 mol Al completely reacts, 2.615 mol Al2O3 can be made.
The balanced chemical equation is: 3H2SO4 + 2Al → Al2(SO4)3 + 3H2. This shows that 3 moles of H2SO4 react with 2 moles of Al. Therefore, using a mole ratio calculation: (18 mol Al) x (3 mol H2SO4 / 2 mol Al) = 27 moles of H2SO4 will react with 18 moles of Al.
In the reaction 4 moles of aluminum will react with 3 moles of oxygen to form 2 moles of aluminum oxide. Since we have 2.0 moles of aluminum, we would need (2.0 mol Al) x (3 mol O2 / 4 mol Al) = 1.5 moles of O2 to react with it.