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How many moles of MgCl2 are there in 313 g of the compound?
Approx 3.29 moles.
If 2Ag2O and 4Ag o2 then if you started with 3.25 moles of Ag2O how many moles of O2 would be produced?
what is ag2o
How many atoms of each element are there in 2Na2CO3?
6. 2 sodium, 1 carbon, and 3 oxygen.
How can you convert from molecules to grams?
Multiply the number of moles by the molecular weight.
What is the reaction between NH4OH and Na2Co3?
The reaction between ammonium hydroxide (NH4OH) and sodium carbonate (Na2CO3) is a double displacement reaction. The products of this reaction are sodium hydroxide (NaOH) and ammonium carbonate ((NH4)2CO3). The balanced chemical equation for this reaction is 2NH4OH + Na2CO3 → 2NaOH + (NH4)2CO3.
How many moles of ions are produced by the dissociation of 0.5 mol of Na2CO3?
When Na2CO3 dissociates, it produces 3 moles of ions: 2 moles of Na+ ions and 1 mole of CO3^2- ions. So, if you have 0.5 moles of Na2CO3, you would produce 1.5 moles of ions in total.
How many moles of Na2CO3 are there in 10.0 L of 2.0 M soluton?
Molarity = moles of solute/Liters of solution Or, for our purposes, Moles of solute = Liters of solution * Molarity Moles Na2CO3 = 10.0 Liters * 2.0 M = 20 moles Na2CO3 --------------------------
What is the number of moles in 8.42g Na2CO3?
It is 0.0794 moles.
If 1 cup is equal to 251.8 grams how many moles of Na2CO3 are used?
To determine how many moles of Na2CO3 are used, you need to know the molar mass of Na2CO3. It is 105.99 g/mol. Then, you can calculate the number of moles using the given mass: mass in grams / molar mass = moles of Na2CO3.
How many moles of washing soda are there in 5g?
Washing soda is sodium carbonate, Na2CO3. Using the atomic weights from the periodic table and the subscripts in the formula, the molar mass of Na2CO3 = 106g/mol. 5g Na2CO3 x (1mol Na2CO3/106g/mol) = 0.05mol Na2CO3
How many grams of sodium carbonate is needed to react with 18.1 ml of 0.23M HCL?
Balanced equation always first!! Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O Find moles HCl by---Molarity = moles of solute/Liters of solution (18.1 ml = 0.0181 L) 0.23 M HCl = moles HCl/0.0181 liters = 0.004163 moles HCl Drive back against sodium carbonate stoichometrically 0.004163 moles HCl (1 mole Na2CO3/2 mole HCl)(105.99 grams/1 mole Na2CO3) = 0.22 grams Na2CO3 needed --------------------------------------------
How many grams of NaOH are formed from reacting 120g sodium carbonate with excess calcium hydroxide?
Write out the equation, and remember to balance each side.Na2CO3 + Ca(OH)2 --> 2NaOH + CaCO3Molecular WeightsNa2CO3: 106 grams/moleNaOH: 40 grams/moleAlways convert your reagents into moles.(120g Na2CO3) x (1 mole Na2CO3/106 grams Na2CO3) = 1.132 molesAccording to the balanced equation, 1 molecule of Na2CO3 generates 2 molecules of NaOH.(1.132 moles Na2CO3) x (2 moles NaOH/1 mole Na2CO3) = 2.264 moles NaOHNow determine the number of grams from 2.264 moles of NaOH.(2.264 moles NaOH) x (40 grams/ 1 mole NaOH) = 90.57 grams NaOH formed.To prevent rounding off too many times, carry out the dimensional analysis in one step:(120g Na2CO3) x (1 mole Na2CO3/106 grams Na2CO3) x(2 moles NaOH/1 mole Na2CO3) x (40 grams/ 1 mole NaOH) = 90.57 grams NaOH
Convert the mass of Na2CO3 to moles with the molecular weight of 105.989?
the answer to your question is 0.0004 g/mol.
How many grams are in 0.577 mol Na2CO3?
To convert moles to grams, you need to multiply the number of moles by the molar mass of the substance. The molar mass of Na2CO3 (sodium carbonate) is 105.99 g/mol. So, to find the grams in 0.577 mol of Na2CO3, you would calculate 0.577 mol * 105.99 g/mol = approximately 60.98 grams of Na2CO3.
Consider this reaction in which 225.4 g of Na2CO3 reacts with 83.6 g of C How many grams of CO will be produced?
To determine how many grams of CO will be produced from the reaction of Na2CO3 with carbon (C), we first need the balanced chemical equation for the reaction, which is: [ \text{Na}_2\text{CO}_3 + 2 \text{C} \rightarrow 2 \text{Na} + 3 \text{CO} + \text{CO}_2 ] Next, we convert the masses of Na2CO3 and C to moles. The molar mass of Na2CO3 is approximately 105.99 g/mol, and for C, it is about 12.01 g/mol. Calculating moles: Moles of Na2CO3 = 225.4 g / 105.99 g/mol ≈ 2.13 moles Moles of C = 83.6 g / 12.01 g/mol ≈ 6.95 moles According to the balanced equation, 1 mole of Na2CO3 reacts with 2 moles of C to produce 3 moles of CO. The limiting reagent here is Na2CO3 since it would require 4.26 moles of C to fully react with 2.13 moles of Na2CO3, which is available (6.95 moles). Therefore, from 2.13 moles of Na2CO3, we can produce: [ 2.13 , \text{moles Na}_2\text{CO}_3 \times \frac{3 , \text{moles CO}}{1 , \text{mole Na}_2\text{CO}_3} = 6.39 , \text{moles CO} ] Finally, converting moles of CO to grams (molar mass of CO is about 28.01 g/mol): [ 6.39 , \text{moles CO} \times 28.01 , \text{g/mol} \approx 178.29 , \text{g CO} ] Thus, approximately 178.29 grams of CO will be produced.
How many moles of ions are produced when 2 mol of Na2CO3 dissociate?
Na2CO32 * 2 = 4 moles sodium.===========================
How many mL of 0.1010 M HCl will be needed to titrate 0.3151 grams of primary standard Na2CO3?
To determine the volume of HCl needed, first calculate the moles of Na2CO3 using its molar mass. Then use the balanced chemical equation to find the moles of HCl required to fully react with the moles of Na2CO3. Finally, use the concentration of the HCl solution to calculate the volume required using the formula: moles = molarity x volume (in liters).
