4 moles
7: Molecules of HCl and KCl have the same number, 1, or chlorine atoms per mole; therefore, if HCl reacts completely with potassium, the same number of moles of potassium chloride as the number of moles of hydrogen chloride present will be formed. (For an ionic compound such as KCl, the term "formula unit" is preferable to "mole", but the mathematical consequences are the same.)
If the reaction is not specified, we can't determine the exact moles of NO formed from NO2 based on this information alone. The reaction and stoichiometry are needed to calculate the moles of NO produced from 8.44 moles of NO2.
The molecular formula for carbonic acid is H2CO3. To find the mass of carbonic acid formed, first calculate the moles of carbon and water. Then, determine the limiting reactant and use it to calculate the moles of carbonic acid formed. Finally, convert the moles of carbonic acid to grams to find the mass.
The most straightforward reaction for the formation of SO3 from SO2 is 2 SO2 + O2 => 2 SO3. If this is the actual reaction for the formation, 3 moles of SO3 are formed from 3 moles of SO2.
To completely replace silver in the solution with copper, you would need an equal number of moles of copper to the moles of silver present. Calculate the moles of silver in the solution using the concentration and volume given. Then use the mole ratio between copper and silver to determine the moles of copper needed, and convert this to grams.
To find the number of moles of Na2CrO4 in 74.3 grams, you need to divide the given mass by the molar mass of Na2CrO4. The molar mass of Na2CrO4 is calculated by adding the atomic masses of all the elements: 2(Na) + Cr + 4(O). Then divide 74.3 grams by the molar mass of Na2CrO4 to get the number of moles.
Oh, dude, let's break it down. So, first off, you gotta calculate the molar mass of Na2CrO4, which is around 205.99 g/mol. Then, you divide the given mass by the molar mass to find the moles. In this case, 74.3 g / 205.99 g/mol gives you approximately 0.36 moles of Na2CrO4. So, there you have it, like, around 0.36 moles of that stuff in 74.3 grams.
300.0 ml of water is added to .40 L of a .400 M Na2CrO4 solution what is the molarity of the resulting solution? Na2CrO4 = 2 Cr +Cr + 4 O's Molar mass = (2*23 + 52 + (4*16) = 162 A .400 M N Na2CrO4 solution has .400 moles of Na2CrO4 in a liter of water. .400 moles of Na2CrO4 = 0.400 * 162 = 64.8 grams of Na2CrO4 in a liter of water. Since you only have .40 L, you have 64.8 grams/liter * 0.4L = 25.92 grams of Na2CrO4 in 0.4 liter of solution. When you add 300.0 ml of water, you have total of 700 ml of solution. You still have 25.92 grams of Na2CrO4, but now you have 700 ml of solution. Molarity = moles of solute per liter of solution. Moles of solute = grams of solute ÷ Molar mass of solute Moles of solute = 25.92 ÷ 162 = 0.16 moles of Na2CrO4. Molarity = 0.16 moles of Na2CrO4 ÷ 0.700 L of solution. Molarity = 0.23 M
Assuming you mean 2.00 M and not 2.00 m, then moles of sodium chromate would be as follows:1.75 L x 2.00 moles/L = 3.50 moles sodium chromate
In the given reaction, the stoichiometry is 1:1 for Fe2O3 to Al2O3. So the number of moles of Al2O3 formed will be the same as the number of moles of Fe2O3 originally present.
The amount of product will be limited by the number of moles of the limitin... reagent.
20.4
5.8
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.
The balanced reaction equation is 4Al + 3O2 -> 2Al2O3. Therefore, 3 moles of O2 reacts with 4 moles of Al to form 2 moles of Al2O3. Since 0.78 mol of O2 is reacted, the number of moles of Al2O3 formed can be calculated using the stoichiometry of the reaction.
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.
7: Molecules of HCl and KCl have the same number, 1, or chlorine atoms per mole; therefore, if HCl reacts completely with potassium, the same number of moles of potassium chloride as the number of moles of hydrogen chloride present will be formed. (For an ionic compound such as KCl, the term "formula unit" is preferable to "mole", but the mathematical consequences are the same.)