To produce 1 mole of urea, 1 mole of carbon dioxide is needed. The molar mass of urea is 60 grams/mol, and the molar mass of carbon dioxide is 44 grams/mol. Therefore, to produce 125 grams of urea, 125 grams/60 grams/mol = 2.08 moles of urea is needed. This means 2.08 moles of carbon dioxide is needed, which is 2.08 moles * 44 grams/mol = 91.52 grams of carbon dioxide needed.
The equation given shows that each formula mass of calcium carbonate produces one formula mass of CO2. The gram formula masses of calcium carbonate and carbon dioxide are 100.09 and 44.01 respectively. Therefore, to produce 4.4 grams of carbon dioxide, 4.4(100.09/44.01), or 10 grams of calcium carbonate, to the justified number of significant digits, are needed.
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.
The amount of carbon dioxide needed to fill a balloon will depend on the size of the balloon. On average, a standard party balloon can hold roughly 0.5 grams of carbon dioxide gas when fully inflated.
To find the grams of carbon dioxide produced, start by calculating the moles of each reactant using their molar masses. Then determine the limiting reactant (the one that forms less product). In this case, oxygen is the limiting reactant. Use the mole ratio from the balanced chemical equation to find the moles of carbon dioxide produced. Finally, convert moles of carbon dioxide to grams using its molar mass.
In 95.0 liters of CO2, there are 3.98 moles of CO2. This is determined from the molar mass of CO2 (44.010 g/mol) and that in 95.0 liters, there is .175 kg. (Density of CO2 is .001842 g/cm3) The stoichiometric equation to produce CO2 from CaCO3 is CaCO3 -> CaO + CO2. Because the equation is balanced with all coefficients being 1, it takes an equivalent number of moles of CaCO3 to produce the CO2. 3.98 moles of CaCO3 = 398 grams. (Molar weight of calcium carbonate is 100.1 g/mol)
22 grams of carbon dioxide contains 12 grams of carbon. This amount of carbon can combine with 32 grams of oxygen to form 44 grams of carbon dioxide.
800 g oxygen are needed.
To calculate the number of moles of carbon dioxide in 19 grams, divide the given mass by the molar mass of carbon dioxide, which is approximately 44 grams/mol. Therefore, 19 grams of carbon dioxide is equal to 19/44 ≈ 0.43 moles.
The equation given shows that each formula mass of calcium carbonate produces one formula mass of CO2. The gram formula masses of calcium carbonate and carbon dioxide are 100.09 and 44.01 respectively. Therefore, to produce 4.4 grams of carbon dioxide, 4.4(100.09/44.01), or 10 grams of calcium carbonate, to the justified number of significant digits, are needed.
11 grams because all is reacted and there is no reactant left over, although if there were only 3 grams of carbon there would have to be 6 grams of oxygen for this to be viable as carbon dioxide is CO2 so the question asked was itself wrong.
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.
The amount of carbon dioxide needed to fill a balloon will depend on the size of the balloon. On average, a standard party balloon can hold roughly 0.5 grams of carbon dioxide gas when fully inflated.
To find the grams of carbon dioxide produced, start by calculating the moles of each reactant using their molar masses. Then determine the limiting reactant (the one that forms less product). In this case, oxygen is the limiting reactant. Use the mole ratio from the balanced chemical equation to find the moles of carbon dioxide produced. Finally, convert moles of carbon dioxide to grams using its molar mass.
In 95.0 liters of CO2, there are 3.98 moles of CO2. This is determined from the molar mass of CO2 (44.010 g/mol) and that in 95.0 liters, there is .175 kg. (Density of CO2 is .001842 g/cm3) The stoichiometric equation to produce CO2 from CaCO3 is CaCO3 -> CaO + CO2. Because the equation is balanced with all coefficients being 1, it takes an equivalent number of moles of CaCO3 to produce the CO2. 3.98 moles of CaCO3 = 398 grams. (Molar weight of calcium carbonate is 100.1 g/mol)
The mass of carbon in carbon dioxide is 12 grams per mole.
A 2-liter bottle of Sprite contains about 5.3 grams of carbon dioxide.
C + O2 = CO2 So the theoretical number of moles are 1 each. number of moles is mass/molecular weight C = 6/12 which is 0.5 O2 = 11/32 which is 0.34375 Oxygen is the limiting reagent. So 0.34375 moles is reacted and this also gives 0.34375 moles CO2 no moles is mass over molecular mass grams is moles x molec mass 0.34375 x 44 = 15.125 grams CO2 formed. learn the technique. this is needed in science