Cryolite, Na_3AlF_6(s), an ore used in the production of aluminum, can be synthe

Question

Cryolite, Na_3AlF_6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. If 15.8 kilograms of Al_2O_3(s), 56 4 kilograms of NaOH(l), and 56.4 kilograms of HF(g) react completely. how many kilograms of cryolite will be produced? Which reactants will be in excess? What is the total mass of the excess reactants left over after the reaction is complete? The first step in the reaction of Alka-Seltzer with stomach acid consists of one mole of sodium bicarbonate (NaHCO_3) reacting with one mole of hydrochloric add (HCl) to produce one mole of carbonic acid (H_2CO_3). and one mole of sodium chloride (NaCl). Using this chemical stoichiometry, determine the number of moles of carbonic add that can be produced from 5 mol of NaHCO_3 and 9 mol of HCl. Which of the two reactants limits the number of moles of H_2CO_3 that can be made? How much excess reactant remains after the reaction?

Explanation / Answer

0.5Al2O3+3NaOH+6HF=Na3AlF6+4.5H2O

number of moles of Al2O3=15.8/102=0.154

similarly NaOH=56.4/40=1.41,HF=56.4/10=5.64

According to stoichiometry1 MOLE of AL2O3 reacts with 6 moles of NaOH and 12moles of HF and form 2moless of cryolite

So 0.154 moles forms 0.924 and 1.848 moles of NaoH and HF

So cryolite formed will be 2*0.154=0.308

So weight of cryolite = 0.308*150=46.2 kg

B)5moles of NaHCO3 forms 5mloes of H2CO3

4moles of excess reactant is H2CO3

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