On complete combustion, 0.246 g of an organic compound gave 0.198 g 0f CO2 and 0.1014 g of H2O. What is the ratio of carbon and hydrogen atoms in the compound?
Given:mass of organic compound --- 0.246 gmass of CO2 formed ----- 0.198 gpercentage of Carbon = 12 x mass of CO2 formed x 100 44 mass of organic compoundpercentage of carbon = 12 /44 x 0.198/0.246 x 100percentage of carbon = 2.376/10.824 x 100% carbon = 0.2195 x 100%carbon = 21.95----------------------------------------------------------21.95% of carbon is present in the compoundPercentage of hydrogen =2/18× mass of H2O/mass of compound×100 = 2×0.1014×100/18×0.246 = 4.58Hence,the composition of compounds areCARBON : 21.95%HYDROGEN : 4.58%.All the best if u guys get more doubts ask me i will help u out of it.Your friend,GRENINJA.
Please help What volume of oxygen gas is needed to react completely with 0.831 L of carbon monoxide gas (CO) to form gaseous carbon dioxide?
What volume of oxygen gas is needed to react completely with 0.831 L of carbon monoxide gas (CO) to form gaseous carbon dioxide? Assume all volume measurements are made at the same temperature and pressure. Answer in units of L.
What volume of oxygen gas is needed to react completely with .538 L of carbon monoxide gas (CO) to form...?
2 CO(g) + O2(g) --> 2 CO2 Use dimensional analysis and equation coefficients to change 0.538 L of carbon monoxide to moles of carbon monoxide, to moles of oxygen, to liters of oxygen: [(0.538 L CO)/1][(1 mole CO)/(22.4 L CO)][(1 mol O2)/(2 mol CO)][(22.4 L O2)/(1 mol O2)] = 0.265 L of O2 Answer: 0.265 liters of oxygen gas is needed to react completely with 0.538 liters of carbon monoxide gas to form gaseous carbon dioxide.
Help! A gas bottle contains 0.650 moles?
a gas bottle contains 0.650 moles of gas at 730 mm Hg pressure. if the final pressure is 1.15 atm, how many moles of gas were added to the bottle. i know the answer is .778 moles but could someone show how they got the answer? thx
What volume of 0.812 M HCl, in milliliters, is required to titrate 1.45 g of NaOH to the equivalence point? ?
the equation is HCl + NaOH ------- NaCl +H2O HCl and NaOH are in the ratio 1:1 mol of NaOH in 1.45g=1.45/40=0.03625mol therefore this is also the mol of HCl required to react with 1.45g of NaOH mol= concentration *volume volume of HCl required=0.03625/0.812=0.0446Litres to ml 0.0446*1000=44.6millilitres
When 0.6g of impure zinc reacted with an excess of hydrochloric acid, 133 mL of hydrogen was collected over water at 10 degrees C. (DETAILS)?
n = PV / RT = (789.6 Torr - 9.21 Torr) x (0.133 L) / ((62.36367 L Torr/K mol) x (10 + 273) K) = 0.00588 mol dry H2 V = nRT / P = (0.005881 mol) x (0.08205746 L atm/K mol) x (298 K) / (1 atm) = 0.144 L H2 at the specified P&T Zn + 2 HCl → ZnCl2 + H2 (0.005881 mol H2) x (1 mol Zn / 1 mol H2) x (65.3820 g Zn/mol) / (0.6 g) = 0.6408 = 64% Zn
Balancing chemical equations and volumes of gases?
These are the easiest kinds of gas problems to solve. from PV = nRT, if P and T are constant, V is proportional to n. 2CO + O2 ---> 2CO2 From the balanced equation you can see half as much O2 is needed as CO, so 0.626L / 2 = 0.313L O2, or More formally, using dimensional analysis 0.626L CO x [1 mole O2 / 2moles CO] = 0.313L O2.
Chemistry Help Please!?
It's always nice when they give / ask for only the moles; it's one less step to do! :) Alright, so you have 3 mols of H2 for every 2 mols of NH3. You also have 1 mol of N2 for every 2 mols of NH3. Knowing the stoichiometry conversion and the mols of both reactants, we can easily figure out the limiting reactant. 0.230 mol N2 (2 mol NH3 / 1 mol N2) = 0.460 mol NH3 produced 0.715 mol H2 (2 mol NH3 / 3 mol H2) = 0.477 mol NH3 produced The limiting reactant is the one that produced the least amount of the product. For this reaction, the limiting reactant is N2. Thus, the amount of moles left of N2 after the reaction takes place is 0 (it was completely used up during the reaction). The reactant in excess is H2. To determine the amount left after the reaction, refer to how many moles were used up during the reaction and change that back into moles of H2. 0.460 mol NH3 (3 mol H2 / 2 mol NH3) = 0.690 mol H2 were used for this reaction 0.715 mol H2 - 0.690 mol H2 = 0.025 mol H2 remain
How many mL of 0.655 M HCl are needed to dissolve 7.85 g of BaCO3?
First, set up a balanced chemical equation. This is an acid + carbonate, so the products are a salt, carbon dioxide, and water. 2HCl + BaCO3 -> BaCl2 + CO2 + H2O Next, find how many moles of BaCO3 7.85g is. The molar mass of BaCO3 is 197.34g/mol, so 7.85g BaCO3 = 0.0398 mol. From our equation, we can see that we need 2 moles of HCl for every 1 mole of BaCO3, so we need 0.0796 mol of HCl. We know that M = mol/L, so 0.655 = 0.0796/x x = 0.122L or 122mL.