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The following reaction is stoichiometric as written$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}+\mathrm{NaOC}_{2} \mathrm{H}_{5} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{NaCl}$but it is often carried out with an excess of $\mathrm{NaOC}_{2} \mathrm{H}_{5}$ to react with any water present in the reaction mixture that might reduce the yield. If the reaction shown was carried out with 6.83 g of $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl},$ how many grams of $\mathrm{NaOC}_{2} \mathrm{H}_{5}$ would beneeded to have a 50 percent molar excess of that reactant?

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$7.53 \mathrm{g}$

Chemistry 101

Chapter 3

Mass Relationships in Chemical Reactions

Chemical reactions and Stoichiometry

Rice University

University of Maryland - University College

University of Kentucky

Lectures

04:02

A chemical reaction is a process that leads to the transformation of one set of chemical substances to another. Chemical reactions can be either spontaneous, requiring no input of energy, or non-spontaneous, typically following the input of some type of energy, such as heat, light or electricity. Chemical reactions are usually characterized by a chemical change, and they yield one or more products after the reaction is complete. Chemical reactions are described with chemical equations, which symbolically present the starting materials, end products, and sometimes intermediate products and reaction conditions. Chemical reactions happen at a characteristic reaction rate at a given temperature and chemical concentration. Typically, reaction rates increase with increasing temperature because there is more thermal energy available to reach the activation energy necessary for breaking bonds between atoms.

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In chemistry, a combination reaction is a chemical reaction in which two or more reactants combine to form more than one product. In a decomposition reaction, one reactant splits into two or more products.

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We have given information for the above question is c: 4, h, 9 c, l plus n, a o c 2, h, 5 and c 4, h, 8 plus c, 2, h, 5, o h, plus n, a c l, the mass of c 4 h, 9 c. L is mass of c 4 h, 9 c. L is equal to 6.83 gram and we need to find the mass of n a o c. 2 is 5 made to have to have a 50 percent, 50 percent molar access axes and first, we will find we will find the number of moles of of moles of c 4 h, 5 c, 4, h, 9 c l, the molar mass of c 4 h, 9 c l is so molar mass of c 4 h, 9 c l is equal to 4 into molar mass of carbon plus 9 into molar. Mass of hydrogen plus molar mass of c l is equal to 4 into molar. Mass of carbon is 12.011 gram per mole plus 9 into molar. Mass of hydrogen is 1.008 gram per mole, plus molar mass of chlorine is 35.453 gram per mole is equal to. We have the calculated molar mass of c 4 h, 9 and c. L is equal to 92.569 gram per mole and the number of moles of moles of c 4 h, 9 c l is moles of numeral moles of c 4 h, 9 c l is equal to given mass of c 4 h, 9 c l divided by molar Mass of c 4 h, 9 c l was equal to given mass of the compound is 6.83 gram. Divided by 92.569 gram per mole is equal to 7.38 multiply 10 raised to the power minus 2 molean 1 mole of c 4 h. L reacts with 1 mole of n a o c 2 h 5, hence 7.838 multiplied and raised to the power minus 2 mole of c 4 h, 9 c l will react with 7.38 multiplied and raised to the power minus 2 mola n, a o c 2 H 5, so here we have since 1 mole of c 4 h 9 cl reacts with 1 mole of a o c. 2 h, 5 point hence 7.38 into 10, raised to the power minus 2 mole of c 4 h, 9 c. L will react, react with 7.38 to 10 raised to the power minus 2 moles of n a o c 2 h 5, since we need we need 50 percent molar mass of n a o c. 2 is 5. We would need meat number of moles of n a o c 2 h. I is equal to 7.38 multiplied and raised to the power minus 2 mole plus 50 percent and to 7.38 into 10 raised to the power. Minus 2 mole is equal to 1111.0 7 point. Tentasto the power minus 2 molthe molar mass of most of n, a o c 2 h 5 is molar mass of n a o c 2 h. 5 is equal to molar mass of sodium plus molar mass of oxygen plus 2 into molar mass of carbon plus 5 into molar. Mass of hydrogen is equal to molar. Mass of sodium is 22.990 gram per mole, plus molar mass of oxygen is 15.999 gram per mole plus 2 into molar. Mass of carbon is 12.011 gram per mole plus 5 into molar. Mass of hydrogen is 1.008 gram per mole is equal to. We have the calculated molar mass of a octo. 5 is 68.051 gram per mole. Therefore, the mass of n, a o c 2 h 5, is needed, is molar mass of n a of c 2 h. 5 is equal to the number of moles of n, a o c 2 h 5 into molar mass of n, a o c 2 h. 5 is equal to number of moles of n. A o c to h. 5 is 11.07 into 10, raise to the power minus 2 mole into molar. Mass of the compound is 68.051 gram per moleis equal to we have calculated mass of na o c a c 2 h. 5 is 7.53. Gram. Is the fin.

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