42 , and 28.
The chemical equation for the combustion of glucose is C6H12O6 + 6O2 -> 6CO2 + 6H2O. From this, we can see that 1 mole of glucose produces 6 moles of CO2. Therefore, 45 grams of glucose (which is approximately 0.25 moles) would produce 6 times that amount of CO2, which is about 90 grams.
The reaction is a limiting reactant problem since we are given the amounts of both reactants. With the excess of sulfur, all the lead will react to form lead sulfide. There won't be any sulfur left over. The analysis of the reaction mixture would show 28.44 g of lead sulfide produced, with excess sulfur not affecting the reaction.
Balanced equation. 2Na + Cl2 >> 2NaCl 46 grams sodium = 2 mol 23 grams Chlorine = 0.65 mol ( I think Chlorine is limiting ) 0.65 mol Cl (2mol Na/1mol Cl ) = 1.3 mol ( you do not have that; Cl limits ) 0.65 mol Cl (2mol NaCl/1mol Cl2 )(58.44g/1mol NaCl ) = 75.9 grams
The balanced chemical equation for the reaction of hydrogen and oxygen to form water is 2H2 + O2 -> 2H2O. Based on the equation, for every 2 grams of hydrogen, 64 grams of oxygen are needed to form 36 grams of water. Thus, if 8 grams of hydrogen react completely with 64 grams of oxygen, the total mass of water formed would be 36 grams.
To find the grams of carbon dioxide produced, first determine the moles of 1-heptanol using its molar mass. Then, use the balanced combustion equation of 1-heptanol to find the moles of CO2 produced. Finally, convert moles of CO2 to grams using its molar mass.
160...cant quite grasp HOW though
266,86 g aluminium chloride are obtained.
The answer is 152 g oxygen.
42,09 g silver chloride are obtained.
To find the mass of H₂O produced, first balance the equation: 2 HCl + Na₂CO₃ → 2 NaCl + H₂O + CO₂. Then calculate the number of moles of HCl in 12.0 grams, use stoichiometry to find the moles of H₂O produced, and finally convert moles of H₂O to grams using the molar mass of water (18.015 g/mol).
First, balance the equation: 4FeS2 + 11O2 → 2Fe2O3 + 8SO2. Calculate the moles of each reactant: moles FeS2 = 1176 g / molar mass of FeS2, moles O2 = 704 g / molar mass of O2. Identify the limiting reactant based on the stoichiometry of the reaction, then use stoichiometry to calculate the grams of Fe2O3 produced.
The amount of product formed in a reaction is determined by the stoichiometry of the reaction. Without the specific reaction given, we can't determine the grams of product produced from 22.4g of a reactant. The balanced chemical equation is needed to calculate the amount of product formed.
The chemical equation for the combustion of glucose is C6H12O6 + 6O2 -> 6CO2 + 6H2O. From this, we can see that 1 mole of glucose produces 6 moles of CO2. Therefore, 45 grams of glucose (which is approximately 0.25 moles) would produce 6 times that amount of CO2, which is about 90 grams.
The atomic mass of hydrogen is 1.008 and the molecular mass of water, with formula H2O, is 18.015. Therefore, the mass of hydrogen to that of water has the ratio of 2(1.008)/18.015 = about 0.1119, and the answer to the problem is 300/0.1119 = 2.68 X 103 grams, to the justified number of significant digits.
To determine the moles of water produced from the reaction of 6.00 grams of propane, first calculate the moles of propane using its molar mass. Then, use the balanced chemical equation to find the moles of water produced based on the stoichiometry of the reaction.
155.2 g
The reaction is a limiting reactant problem since we are given the amounts of both reactants. With the excess of sulfur, all the lead will react to form lead sulfide. There won't be any sulfur left over. The analysis of the reaction mixture would show 28.44 g of lead sulfide produced, with excess sulfur not affecting the reaction.