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What is g of NaHCO3?

Updated: 12/4/2022
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Related questions

How many moles of NaHCO3 are present in a 6.10 g sample?

Divide 6.10 (g NaHCO3) by 84.007 (g.mol−1 NaHCO3) to get 0.0726 mol NaHCO3


NaHCO3 is the active ingredient in baking soda. How many grams of oxygen are in 0.53 g of NaHCO3?

The mass of oxygen is o,303 g.


What is the product for NaHCO3 and CaHPO4?

NaHCO3(aq) + CaHPO4(aq) → NaCaPO4(aq) + CO2(g) + H2O(l)


How many moles are in 2.10 kg of NaHCO3?

For this you need the atomic (molecular) mass of NaHCO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaHCO3=84.0 grams110 grams NaHCO3 / (84.0 grams) = 1.31 moles NaHCO3


What mass of NaHCO3 would be needed to neutralize a spill consisting of 12.5 mL of 5.0 M HCl?

12.5 mL * 5.0 (m)mol/(m)L HCl = 62.5 mmol spilled HClneeds62.5 mmol NaHCO3 = 62.5 mmol * 84.01 (m)g/(m)mol NaHCO3 = 5250 mg NaHCO3 = 5.25 g pure NaHCO3


If 2.10 g of NaHCO3 are decomposed then how many grams of water are produced?

NaHCO3 ----> H2O Mass 2.10g 0.045g RAM 84 g/moles 18 g/moles number of moles 0.025moles 0.025moles


What is the percent CO2 in sodium bicarbonate?

Sodium Bicarbonate come from absorption of CO2 into NaOH Total reaction is 2NaOH + CO2 -> NaHCO3 + H2O Each mole of NaHCO3 come from absorption of 1 mole CO2 Molecular weight of NaHCO3 is 84 g/mol and CO2 is 44 g/mol It is that 84 g of NaHCO3 had 44 g of CO2 By weight ratio it is 52% CO2 in Sodium Bicarbonate.


How many moles of NaHCO3 are present in 3.00 g sample?

3.00 grams NaHCO3 (1 mole/84.008 grams) = 0.0357 moles of sodium bicarbonate


How many grams are in 1 mol nahco3?

1 mole of NaHCO3 Na = 1 * 22.99 g = 22.99 g H = 1 * 1.01 g = 1.01 g C = 1 * 12.01 g = 12.01 g O = 3 * 16.00 g = 48.00 g Total = 84.01 g There are 84.01 grams in one mole of NaHCO3.


How many moles of hcl are needed to react with 0.35 g of NAHCO3?

The answer to the conversion is that 35.0 grams of hydrochloride (HCL) equals 0.76 moles. The conversion rate is 35.0 grams divided by 46 gram per mole. A mole is the molecular weight of a substance.


How much sodium bicarbonate would be required to prepare 500ml of a 100mM solution?

500 mL * 100(mMol/mL) = 50 mMol NaHCO3 , hence50 mMol NaHCO3 = 50(mMol) * 84(mg/mMol) = 4200 mg = 4.2 g NaHCO3 in 500 mL


If a typical antacid tablet contains 2 grams of sodium hydrogen carbonate how many moles of carbon dioxide should one tablet yield?

NaHCO3 = 84.007 g/mol Figure out the mole fraction = 2g / 84.007 g/mol The unit of the answer is mol, which in this context is the same as mole. Since you only have one carbon in NaHCO3 you cannot have more moles of CO2 than moles of NaHCO3.