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The equation for the reaction is 4 Na + O2 -> 2 Na2O. This shows that, for complete reaction, one mole of oxygen is required for each four gram atomic masses of sodium. The gram atomic mass of sodium is 22.9898; therefore, 46 grams of sodium constitutes 2.00 moles of sodium, to more than the justified number of significant digits. The gram molecular mass of diatomic oxygen is 31.9988; therefore 160 grams of oxygen constitutes 5.000 moles of diatomic oxygen, to more than the justified number of significant digits. This is well over the minimum amount of oxygen required for complete reaction of all the sodium present. Each two gram atomic masses of sodium produces one gram formula mass of sodium oxide; therefore, the number of gram formula masses of sodium oxide produced is 1.00, to at least the justified number of significant digits.
This amount may be different because rust is not a clearly definite compound.
160mm is 6.29921 inches. Direct Conversion Formula 160 mm* 1 in 25.4 mm = 6.299212598 in
160 kg = 352 lbs The formula to convert kg to lbs 160 kg*2.2046 lbs 1 kg=352.7396195 lbs
160 grams
That is the chemical formula. The name would be iron (III) oxide, or ferric oxide in the old system.
The answer is C10H14N2
The balanced reaction equation is 2Fe2O3 + 3C = 4Fe + 3CO2 Note the molar ratios are 2:3::4:2 Next calculate the acutal moles of Fe2O3 First using the Periodic Table Calculate the Mr ( Relative molecular mass) of Fe2O3 Fe x 2 = 56 x 2 = 112 O x 3 = 16 x 3 = 48 112 + 48 = 160 mol(Fe2O3) = 16.5 / 160 = 0.103125 Equivlanet to ;2; moles. mol(C) = 0.103125 x 3/2 = 0.1546875 equivalent to '3 moles. Mass(C) = 0.154687 x 12 = 1.85625 g (The Answer!!!!)
Good one. Fe2O3 is 160g/mol, so %comp of Fe is 112/160=0.7 or 70%. 1x106g Fe/70=x/30, so x=428,571g O2, so mass of Fe2O3 is 1,428,571g. 1,428,571/78=x/22, so x=402,930g of other rock, so mass of ore deposit = 1,428,571+402,930=1,831,501g=1,831.5kg.
160...cant quite grasp HOW though
6g hydrogen would be required for 160g ferric oxide in this reaction. The relative atomic weights of the elements are: Hydrogen - 1 Oxygen - 16 Iron - 56 giving the relative atomic weights of the compounds (on the left of the equation): Fe2O3 = 56×2 + 16×3 = 160 3H2 = 3×(1×2) = 6 So for every 160 units of mass of iron III oxide there will be 6 units of mass of hydrogen required. → for 160g of iron III oxide ÷ 160 × 6 = 6 g of hydrogen.
The equation for the reaction is 4 Na + O2 -> 2 Na2O. This shows that, for complete reaction, one mole of oxygen is required for each four gram atomic masses of sodium. The gram atomic mass of sodium is 22.9898; therefore, 46 grams of sodium constitutes 2.00 moles of sodium, to more than the justified number of significant digits. The gram molecular mass of diatomic oxygen is 31.9988; therefore 160 grams of oxygen constitutes 5.000 moles of diatomic oxygen, to more than the justified number of significant digits. This is well over the minimum amount of oxygen required for complete reaction of all the sodium present. Each two gram atomic masses of sodium produces one gram formula mass of sodium oxide; therefore, the number of gram formula masses of sodium oxide produced is 1.00, to at least the justified number of significant digits.
V (VOLUME) = M (MASS) / D (DENSITY) All you have to do is to find the density of methylated spirit and use the formula
There are several different possible reactions of Fe2O3 with CO, depending on temperature and ratio of reactants. The simplest is probably Fe2O3 + CO ==>2FeO + CO21.00 Kg x 1000 g/Kg x 1 mole Fe2O3/160 g = 6.25 moles Fe2O3 moles CO2 produced = 6.25 moles CO2 Volume CO2 at STP = 6.25 moles x 22.4 L/mole = 140 Liters
This amount may be different because rust is not a clearly definite compound.
The percentage can be calculated using the formula (152/160) * 100 = 95%. Therefore, 152 is 95% of 160.
72.575 kg Algebraic Steps / Dimensional Analysis Formula 160 lb*1 kg 2.2046 lb=72.5747792 kg