For this you need the atomic (molecular) mass of CaCO3. Take the number of moles and multiply it by the Atomic Mass. Divide by one mole for units to cancel. CaCO3= 100.1 grams
2.50 moles CaCO3 × (100.1 grams) = 250.25 grams CaCO3
1. Calculate formula massCalcium carbonate has chemical formula CaCO3.Its formula mass is 40.1 + 12.0 + 3(16.0) = 100.12. Apply formula to calculate number of moles of CaCO3Amount of CaCO3= mass/formula mass= 50/100.1= 0.50mol
Assuming each Tums tablet contains 500 mg of calcium carbonate, there is a total of 6 grams of calcium carbonate in one roll of Tums (12 tablets x 500 mg). To calculate the number of moles, divide the mass by the molar mass of calcium carbonate (100.09 g/mol), yielding around 0.06 moles of calcium carbonate in one roll of Tums.
To calculate the number of moles of stomach acid neutralized by calcium carbonate, you first need to convert the mass of calcium carbonate (600 mg) to grams (0.6 g). Then, use the molar mass of calcium carbonate (100.09 g/mol) to find the number of moles. Finally, use the balanced chemical equation to determine the moles of stomach acid neutralized.
The molar ratio between calcium carbonate and calcium oxide is 1:1. So, 25 moles of calcium carbonate will produce 25 moles of calcium oxide. The molar mass of calcium oxide is 56.08 g/mol, so the mass of calcium oxide produced will be 25 moles * 56.08 g/mol = 1402 g.
To answer this we must first find the molar mass of calcium carbonate. CaCO3Ca= 40.08gC=12.01gO= 16.00g (we have three oxygens so 16.00x3 is 48.00g)40.08+12.01+48.00= 100.09 gNow that we have the molar mass we can find how many grams there are:1.25 moles CaCo3 x (100.09 g CaCO3/ 1 mole CaCO3)= 125.11 grams CaCO3Therefore we'd have about 125 grams of CaCO3
2,8 moles of calcium carbonate have 240,208 g.
1. Calculate formula massCalcium carbonate has chemical formula CaCO3.Its formula mass is 40.1 + 12.0 + 3(16.0) = 100.12. Apply formula to calculate number of moles of CaCO3Amount of CaCO3= mass/formula mass= 50/100.1= 0.50mol
Assuming each Tums tablet contains 500 mg of calcium carbonate, there is a total of 6 grams of calcium carbonate in one roll of Tums (12 tablets x 500 mg). To calculate the number of moles, divide the mass by the molar mass of calcium carbonate (100.09 g/mol), yielding around 0.06 moles of calcium carbonate in one roll of Tums.
To find the amount of calcium chloride produced, first calculate the moles of calcium carbonate and hydrochloric acid using their molar masses. Then, determine the limiting reactant and use stoichiometry to find the moles of calcium chloride produced. Finally, convert moles of calcium chloride to grams using its molar mass.
To calculate the grams of calcium carbonate needed, you first need to determine the moles of carbon dioxide produced (using the ideal gas law). Then, since 1 mole of CO2 is produced for every mole of CaCO3 consumed, you can convert moles of CO2 to moles of CaCO3. Finally, use the molar mass of CaCO3 to convert moles to grams.
To calculate the number of moles of stomach acid neutralized by calcium carbonate, you first need to convert the mass of calcium carbonate (600 mg) to grams (0.6 g). Then, use the molar mass of calcium carbonate (100.09 g/mol) to find the number of moles. Finally, use the balanced chemical equation to determine the moles of stomach acid neutralized.
Calcium carbonate, CaCO3 has formula mass of 40.1+12.0+3(16.0) = 100.1Amount of CaCO3 = 1.719/100.1 = 0.0172molThere are 0.0172 moles of calcium carbonate in a 1.719 gram pure sample.
The molar mass of calcium carbonate (CaCO3) is 100.09 g/mol. Therefore, the mass of 0.5 moles of calcium carbonate would be 0.5 mol x 100.09 g/mol = 50.045 grams.
To convert moles to grams, you need to use the molar mass of calcium carbonate (CaCO3). The molar mass of CaCO3 is approximately 100.1 g/mol. To calculate the grams in 2.38 moles of CaCO3, you would multiply the number of moles (2.38) by the molar mass (100.1 g/mol), which gives you approximately 238 grams.
For a partly ionically bonded compound such as calcium carbonate, the gram formula mass is substituted for a mole, which technically exists only for purely covalently bonded compounds. The gram formula mass for calcium carbonate is 100.09. Therefore, 200 grams constitutes 200/100.09 or 2.00 gram formula masses of calcium carbonate, to the justified number of significant digits.
The molar ratio between calcium carbonate and calcium oxide is 1:1. So, 25 moles of calcium carbonate will produce 25 moles of calcium oxide. The molar mass of calcium oxide is 56.08 g/mol, so the mass of calcium oxide produced will be 25 moles * 56.08 g/mol = 1402 g.
To answer this we must first find the molar mass of calcium carbonate. CaCO3Ca= 40.08gC=12.01gO= 16.00g (we have three oxygens so 16.00x3 is 48.00g)40.08+12.01+48.00= 100.09 gNow that we have the molar mass we can find how many grams there are:1.25 moles CaCo3 x (100.09 g CaCO3/ 1 mole CaCO3)= 125.11 grams CaCO3Therefore we'd have about 125 grams of CaCO3