How many liters of H2 gas at STP can be produced by the reaction of 4.60 g of Na and excess water, according?
First divide by the molar mass of Na to find out how many moles of Na you have. Second, 2 moles Na produces 1 mole H2 (1/2 as much). So take your answer above and divide it by 2 to get moles H2 produced Third, Multiply moles H2 produced by 22.4Litres per mole at STP to find the volume of H2 gas produced.
How much H2 gas at STP can be produced by the reaction 2 Na(s)+2 H2O(l) to H2(g)+2 NaOH(aq)......?
You have 2.20g of Na and excess water, so that makes the Na the Limiting Reactant. Use Dimensional Analysis to find out how many moles of H2 can be made. If you look at the balanced equation, you will see that 2 moles of Na react to make 1 mole of H2 (you will need this info below) 2.20g Na x 1mol Na/22.99g Na x 1mol H2/2mol Na =.0478 mol H2 Now make use of PV = nRT and solve for V Rearranged you get: V = nRT/P The reaction is ran at STP, so that means pressure is 1 atm and the temperature is 273K V = (.0478)(.0821)(273)/1 = 1.07 L
How much h2 gas at stp can be produced by the reaction...?
2.20g Na x 1mol Na/22.99g Na x 1mol H2/2mol Na =0.0478 mol H2 PV = nRT solve for V: V = nRT/P V = (0.0478)(0.0821)(273K)/1 = 1.07 L
What mass of H2 will react with 22.4l of O2 at STP to produce 36 grams of water?
This question has more information than you need to answer but…At STP, one mole of any gas (ideal, of course) will occupy 22.4 L so we have 1 mole of oxygen (O2). You can calculate the number of moles, if you wish withPV=nRT or, for n, n= PV/RTAt STP, P = 1 atm, T = 273 K, R is the gas constant (=0.082 atm-L/mol-K) and we are given the V = 22.4. So, the number of moles is:n= PV/RT = (1 * 22.4)/(0.082 * 273) = 1 mole.The reaction in question is:2 H2 + O2 → 2 H2OSo, you need twice as much of H2 as O2 or 2 moles of H2. Hydrogen molecular weight is 2.016 g/mole and you need two moles or 4.032 g.Note that 1 mole of oxygen (O2) has a mass of 32 g so when you add the hydrogen to get water the total mass should be 32+4 or 36 g (rounding things off).
How much H2 gas at STP can be produced by...?
Work out the moles of Na that you have started with moles = mass / molar mass molar mass Na = 22.99 g/mol moles Na = 2.50 g / 22.99 g/mol = 0.1087 moles Na Now work out how many moles of H2 you will get form this many moles of Na The balanced equation 2Na(s) + 2H2O(ℓ) −! H2(g) + 2NaOH(aq) tells you that 2 moles Na react to produce 1 mole H2 Therefore 1 mole Na will react to give 1/2 mole H2 So 0.1087 moles Na will give (1/2 x 0.1087) moles H2 = 0.0544 moles (3 sig figs) You can produce 0.0544 moles of H2 from 2.50 g of Na Now, 1 mole of any ideal gas occupies a volume of 22.4 L at STP STP is standard temp and pressure (0 deg C and 1 atm) So if you know 1 mole has a voluem of 22.4 L then you can calculate the volume of gas by multiplying moles by 22.4L (ONLY if at STP) V = 22.4 L x 0.0544 moles V = 1.22 L (3 sig figs)
How many liters of H2 gas at STP can be produced by the reaction of 4.60gNa and excess water, according to the?
From the balanced equation: 2Na + 2H2O → 2NaOH + H2 2 mol Na will produce 1 mol H2 Molar mass Na = 23g/mol 4.60 g = 4.60/23 = 0.2 mol Na This will produce 0.1 mol H2 At STP , 1 mol H2 has volume 22.4L At STP 0.1 mol has volume 22.4*0.1 = 2.24L That is the volume of H2 produced
How man liters of H2 gas at STP can be produced by the reaction of 4.60g of Na and excess water?
Here's how to do it: change grams to moles to moles to liters 4.60 g Na * (1 mol Na / 23g Na) * (1 mol H2 / 2 mol Na) * (22.4 L H2 / 1 mol H2) = 2.24 L H2
What is the reaction of Al + NaOH?
Al will react with aqueous NaOH to give Sodium Aluminate with rapid evolution of H2 gas.2 Al + 2 NaOH + 2 H2O → 2 NaAlO2 + 3 H2But this is just the theoretical way to write the reaction. In aqueous reaction conditions NaAlO2 will be accompanied by 2H2O i.e. “NaAlO2.2H2O”.This actually exists as Al[OH4]- ions i.e. Sodium tetrahydroxoaluminate(III) ions. Thus the reaction will be -2Al (s) + 2NaOH(aq) + 6H2O(l) → 2 Na+ [Al(OH)4] –(aq) + 3H2(g)Even going further -The structure of [Al(OH)4] – is complicated than it appears to be. Its structure changes with both pH and concentration.Between pH 8 - 12 the ions polymerise using OH bridges and each of the Al is octahedrally coordinated. Roughly like this -2. In dilute solution above pH values 13, a tetrahedral [Al(OH)4] – ion exists.3. In concentrated solution above 1.5 M and at pH values greater than 13 the ion exists as a dimer. [(OH)3 Al - O - Al (OH)3] 2-.I have taken these values from Concise Inorganic Chemistry by J. D. Lee. Image Courtesy : Google!Hope this helps!
How many liters of H2 are needed to produce 2.50 L of CH4 at STP?
[math]\require{cancel}[/math]The first step is to find the balanced chemical equation for this reaction. Let’s assume hydrogen gas ([math]\text{H}_2[/math]) reacts with carbon dioxide ([math]\text{CO}_2[/math]) to produce methane ([math]\text{CH}_4[/math]) and a byproduct of water ([math]\text{H}_2\text{O}[/math]) at STP. It doesn’t, but let’s assume. We start off by writing the reactants and products in the skeleton equation.[math]\text{CO}_2\text{(g)}+\text{H}_2\text{(g)}\longrightarrow \text{CH}_4\text{(g)}+\text{H}_2\text{O(l)}[/math]Next, we balance the equation. Hopefully, you already know how to do this, so I’ll just do it for you.[math]\text{CO}_2\text{(g)}+4\text{H}_2\text{(g)}\longrightarrow \text{CH}_4\text{(g)}+2\text{H}_2\text{O(l)}[/math]Hydrogen gas and methane are both gases at STP, so we can use the conversion factor of [math]\frac{22.4\text{ L}}{1.00\text{ mol}}[/math] to convert from moles to volume and vice-versa.We also know that 4 moles of hydrogen gas react to produce 1 mole of methane, so our other conversion factor is [math]\frac{4.00\text{ mol H}_2}{1.00\text{ mol CH}_4}[/math].Finally, we set up dimensional analysis (or whatever you want to call it) and cancel units to reach our answer.[math]\begin{array}{c|c} 2.50\text{ }\cancel{\text{L}}\text{ }\cancel{\text{CH}_4} & 1.00\text{ }\cancel{\text{mol}} & 4.00\text{ }\cancel{\text{mol}}\text{ H}_2 & 22.4\text{ L} \\ \hline & 22.4\text{ }\cancel{\text{L}} & 1.00\text{ }\cancel{\text{mol}}\text{ }\cancel{\text{CH}_4} & 1.00\text{ }\cancel{\text{mol}} \end{array}[/math][math]\implies 10.0\text{ L H}_2[/math]By the way, this is known as the Sabatier reaction.It involves the reaction of hydrogen with carbon dioxide at elevated temperatures (optimally 300–400 °C) and pressures in the presence of a nickel catalyst to produce methane and water. Optionally, ruthenium on alumina (aluminium oxide) makes a more efficient catalyst.Sabatier reaction - Wikipedia