This reaction is:
2 Al + 2 H2O + 2 NaOH = 2 NaAlO2 + 3 H2
From 4 moles of Al 6 moles of hydrogen are obtained.
4 moles also 2 moles of oxygen will be consumed
1. Write the balanced equation: 2Al(s) + 6HCl(aq) ==> 2AlCl3(aq) + 3H2(g)2. Convert 3.70 g Al to moles of Al: 3.70 g x 1 mol/26.98 g 3. Use stoichiometric ratios in balanced equation to find moles H2 produced: answer from step 2 (moles Al) x 3 moles H2/2 moles Al = moles H2 produced 4. Convert moles H2 produced found in step 3 to grams H2 using molar mass of H2
First we are going to find number of H2 moles form it mass H2 moles=(2.37*10^-4)/2 = 1.185*10^-4moles Since 3 moles of H2 makes 2 moles of NH3 then by using this ration, we can find number of moles in NH3 NH3 moles= (2 * 1.185*10^-4)/3 = 0.79*10^-4 moles Finally, we find number of NH3 molecules by multiplying the number of moles with (6.022*10^23). #NH3 molecules = (0.79*10^-4) *(6.022*10^23) =4.75 * 10^19 molecules of NH3 Good luck Enas
Balanced equation first.2H2 + O2 --> 2H2OGet moles products.4 grams H2 (1 mole H2/2.016 grams) = 1.984 moles H264 grams O2 (1 mole O2/32 grams) = 2.000 moles O2I suspect hydrogen gas of limiting and driving the reaction.1.984 moles H2 (1 mole O2/2 moles H2) = 0.992 moles O2 ( you have more than this in equation )2.000 moles O2 (2 mole H2/1 mole O2) = 4.000 moles H2 ( you do not have this much and H2 will drive this reaction )1.984 moles H2 (2 moles H2O/2 moles H2)(18.016 grams/1 mole H2O)= 36 grams water produced====================
It depends on what you are reacting the sodium with to generate hydrogen gas. The question is incomplete and cannot be answered as it is written
4 moles also 2 moles of oxygen will be consumed
1. Write the balanced equation: 2Al(s) + 6HCl(aq) ==> 2AlCl3(aq) + 3H2(g)2. Convert 3.70 g Al to moles of Al: 3.70 g x 1 mol/26.98 g 3. Use stoichiometric ratios in balanced equation to find moles H2 produced: answer from step 2 (moles Al) x 3 moles H2/2 moles Al = moles H2 produced 4. Convert moles H2 produced found in step 3 to grams H2 using molar mass of H2
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 =========================
First we are going to find number of H2 moles form it mass H2 moles=(2.37*10^-4)/2 = 1.185*10^-4moles Since 3 moles of H2 makes 2 moles of NH3 then by using this ration, we can find number of moles in NH3 NH3 moles= (2 * 1.185*10^-4)/3 = 0.79*10^-4 moles Finally, we find number of NH3 molecules by multiplying the number of moles with (6.022*10^23). #NH3 molecules = (0.79*10^-4) *(6.022*10^23) =4.75 * 10^19 molecules of NH3 Good luck Enas
8.086g
Balanced equation first.2H2 + O2 --> 2H2OGet moles products.4 grams H2 (1 mole H2/2.016 grams) = 1.984 moles H264 grams O2 (1 mole O2/32 grams) = 2.000 moles O2I suspect hydrogen gas of limiting and driving the reaction.1.984 moles H2 (1 mole O2/2 moles H2) = 0.992 moles O2 ( you have more than this in equation )2.000 moles O2 (2 mole H2/1 mole O2) = 4.000 moles H2 ( you do not have this much and H2 will drive this reaction )1.984 moles H2 (2 moles H2O/2 moles H2)(18.016 grams/1 mole H2O)= 36 grams water produced====================
2 mol H2 x 2 mol H2O/2 mol H2 x 18 g H2O/1 mol H2O = 36 g H2O
It depends on what you are reacting the sodium with to generate hydrogen gas. The question is incomplete and cannot be answered as it is written
The balanced chemical equation for the reaction between aluminum (Al) and oxygen (Oβ) to form aluminum oxide (AlβOβ) 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 (AlβOβ). 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.
depends on how much aluminum oxide you have, 1gram, 2 billion Kg? how many? cant find the number of moles of oxygen without knowing the mass of the al203
Four:2 Al + 3 Cl2 --> 2 AlCl3so: 4 Al + 6 Cl2 --> 4 AlCl3
4. The chemical equation is Cu + H2SO4 → CuSO4 + H2. Copper and sulfuric acid have the same coefficient (1), so the same number of moles of copper and sulfuric acid are used.