Stoichiometry molar ratios from balanced equations

We need to calculate the amount of methyl tert-bu tyl ether that could theoretically be produced from 26.3 g of isobutylene and compare that theoretical amount to the actual amount (32.8 g). As always, stoichiometry problems begin by calculating the molar masses of reactants and products. Coefficients of the balanced equation then tell mole ratios, and molar masses act as conversion factors between moles and masses. 

Compare molar amounts using stoichiometry ratios from the balanced equation. Solve for the unknown molar amount. 

To see how molarity can be used in equation stoichiometry problems, let s take a look at the thought process for calculating the number of milliliters of 1.00 M AgN03 necessary to precipitate the phosphate from 25.00 mL of0.500 M Na3P04. The problem asks us to convert from amount of one substance in a chemical reaction to amount of another substance in the reaction, so we know it is an equation stoichiometry problem. The core of our setup will be the conversion factor for changing moles of sodium phosphate to moles of silver nitrate. To construct it, we need to know the molar ratio of AgN03 to Na3P04, which comes from the balanced equation for the reaction. 

In Chapters 3 and 4, we encountered many reactions that involved gases as reactants (e.g., combustion with O2) or as products (e.g., a metal displacing H2 from acid). From the balanced equation, we used stoichiometrically equivalent molar ratios to calculate the amounts (moles) of reactants and products and converted these quantities into masses, numbers of molecules, or solution volumes (see Figure 3.10). Figure 5.11 shows how you can expand your problem-solving repertoire by using the ideal gas law to convert between gas variables (F, T, and V) and amounts (moles) of gaseous reactants and products. In effect, you combine a gas law problem with a stoichiometry problem it is more realistic to measure the volume, pressure, and temperature of a gas than its mass. 

A titration problem is an applied stoichiometry problem, so we will need a balanced chemical equation. We know the molarity and volume for the NaOH solution, so we can find the number of moles reacting. The mole ratio from the balanced equation lets us calculate moles of H2SO4 firom moles of NaOH. Because we know the volume of the original H2SO4 solution, we can find its molarity. 

Two things must be considered. First is the stoichiometry. We need to look at the chemical equations to see the molar ratios far the production of hydrogen far each of the metals. Then we need to look at the molar mass of each metal. Those with a larger molar mass will tend to need more mass than those with smaller molar masses. Ta make the decision, it would make sense to choose some mass, say 100 g, far each metal, convert that mass to moles, and use the mole-to-mole ratio far the production of hydrogen (from each balanced chemical equation) to compare the amount of hydrogen produced. 

To do this problem, first determine the amount of MgC03 present (by multiplying 670 kg by 0.277) and then do a stoichiometry problem with this mass. Convert to moles of MgC03 using the molar mass, use the mole-to-nnole ratio from the balanced chemical equation to convert moles of MgC03 to moles of CO2, and use the molar mass of CO2 to get the mass. 

In a balanced equation, the number of moles of one substance is stoichiometri-cally equivalent to the number of moles of any other substance. The term stoi-chiometrically equivalent means that a definite amount of one substance is formed from, produces, or reacts with a definite amount of the other. These quantitative relationships are expressed as stoichiometricolly equivalent molar ratios that we use as conversion factors to calculate these amounts. Table 3.3 presents the quantitative information contained in the equation for the combustion of propane, a hydrocarbon fuel used in cooking and water heating ... 

As in any stoichiometry problem we need a balanced chemical equation, so we will start by writing the half-reaction for gold reduction. To determine the mass of gold deposited, we must calculate the number of moles of electrons used from the current and the time. We can use the half-reaction to obtain a mole ratio and convert moles of electrons into moles of gold. Once we have moles of gold, we convert to mass using the molar mass, as we have done many times in stoichiometry problems. 

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