Stoichiometry problems
To find the empirical formula, we need to determine the ratio of each element in the compound. First, find the moles of each element: K = 0.104 mol C = 0.052 mol O = 0.156 mol Next, divide each mole value by the smallest mole value to get the ratio: K = 0.104 mol / 0.052 mol = 2 C = 0.052 mol / 0.052 mol = 1 O = 0.156 mol / 0.052 mol = 3 Therefore, the empirical formula is K2CO3.
The ratio of effusion rates for two gases is given by the square root of the inverse ratio of their molar masses. The molar mass of Ar is approximately 40 g/mol, and for Kr it is approximately 84 g/mol. So, the ratio of effusion rates for Ar and Kr is √(84/40) ≈ 1.3.
You use the Mole-to-Mole ratio. If the equation is 2CH4 + 2H2O = 6H2 + 2CO, then you would start with your given, 8.0 mol CO and multiply that with your mol-to-mol ratio which is (2mol CO/ 2 mol CH4). Your answer will be 8.0 mol.
The molar ratio of hydrogen to oxygen in the compound is 1:1. This means the compound is water (H2O), which has a molecular mass of 18.0 g/mol, not 34.0 g/mol. The given molecular mass of 34.0 g/mol does not match the properties of water.
Fe2O3 + 2Al --> 2Fe + Al2O3Before:50.0g + 50.0g > 0.0g + (not important)159.69(g/mol) + 26.98(g/mol)In mol (before reaction):+0.3131 mol + 1.853 mol (excess)Reaction (used reactant > formed Fe):-0.3131 mol - 0.6262 mol > + 0.6262 mol FeRemaining (= before - used):0.0 mol Fe2O3 + 1.227 mol Al > 0.6262 mol Fe, this should be multiplied by the molar mass of Fe to get mass in grams: 0.6262 (molFe) * 55.85 (g/molFe) = 34.97 = 35.0 g Fe
To find the empirical formula, we need to determine the ratio of each element in the compound. First, find the moles of each element: K = 0.104 mol C = 0.052 mol O = 0.156 mol Next, divide each mole value by the smallest mole value to get the ratio: K = 0.104 mol / 0.052 mol = 2 C = 0.052 mol / 0.052 mol = 1 O = 0.156 mol / 0.052 mol = 3 Therefore, the empirical formula is K2CO3.
The mole ratio of BaCl2 to AgCl is 1:2. This means that for every 1 mole of BaCl2, 2 moles of AgCl are produced in the chemical reaction.
The ratio of effusion rates for two gases is given by the square root of the inverse ratio of their molar masses. The molar mass of Ar is approximately 40 g/mol, and for Kr it is approximately 84 g/mol. So, the ratio of effusion rates for Ar and Kr is √(84/40) ≈ 1.3.
The ratio of moles of solute to liters of solution is called molarity. Molarity is expressed in moles per liter (mol/L) and is commonly used to quantify the concentration of a solute in a solution.
In 1.5 mol of SnCl2, there would be 1.5 mol of Sn2+ ions and three times as many Cl- ions due to the 1:2 ratio of SnCl2, so there would be a total of 4.5 mol of ions present.
The balanced chemical equation for the reaction between Mg and Cl2 to form MgCl2 is: Mg + Cl2 -> MgCl2 From the equation, it can be seen that 1 mol of Mg reacts with 1 mol of Cl2 to produce 1 mol of MgCl2. Therefore, the ratio of Cl2 to Mg in MgCl2 would be 1:1.
The molar mass of calcium is 40.08 g/mol and chlorine is 35.45 g/mol. The number of moles of calcium is 3.609g / 40.08 g/mol = 0.09 mol, and for chlorine it is 6.384g / 35.45 g/mol = 0.18 mol. The ratio of moles is 0.09 mol Ca to 0.18 mol Cl, thus the empirical formula is CaCl2.
Concentration = Molarity = mol/L24 g NaCl = ?? mol NaCl?? mol NaCl/2 L water = ?? M (M is unit of molarity)
The molar mass of Cu is 63.55 g/mol and of O is 16 g/mol. The moles of Cu used is 127g / 63.55 g/mol = 2 mol, and the moles of O used is 32g / 16 g/mol = 2 mol. Since the mole ratio of Cu to O in CuO is 1:1, then 2 mol of Cu will react with 2 mol of O to form 2 mol of CuO. The mass of 2 mol of CuO is 159.6g.
To find the empirical formula, calculate the moles of iron and chlorine in the compound. Then, determine the ratio of moles of iron to moles of chlorine. The mole ratio is 1:2, so the empirical formula is FeCl2.
You use the Mole-to-Mole ratio. If the equation is 2CH4 + 2H2O = 6H2 + 2CO, then you would start with your given, 8.0 mol CO and multiply that with your mol-to-mol ratio which is (2mol CO/ 2 mol CH4). Your answer will be 8.0 mol.
Wilhelm Ostwald used the first the word mol in 1894.