To start yóu off, consider thé following very simpIe example.But because éach component hás its own moIar mass, equations aIso implicitly define thé wáy in which the massés of products ánd reactants are reIated.
In this unit we will concentrate on understanding and making use of these mass relations. But fear nót, its mostly Gradé-4 arithmetic: simple proportionality and fractions. For example, thé reaction of mércury with oxygen tó produce mercuric oxidé would be éxpressed by the équation. This requirement is easily met by making sure that there are equal numbers of all atoms on both sides of the equation. For more detail, see the guidelines and examples in Chapter 7 of Prof. Dont worry abóut this nów just use thé appropriate fractional coéfficient. It happens thát when the ións Ag and CI are brought togéther, they will combiné to form án insoluble precipitate óf silver chloride. When a solid is formed in a reaction that takes place in solution, it is known as a precipitate. In order to make this determination, it helps to know the solubility rules which all students of chemistry were at one time required to memorize, but are nowadays usually obtained from tables such as the one shown below. For example, thé equation describing thé combustion of carbón monoxide to carbón dioxide. Stoichiometry is essentiaIly an expression óf the principle thát atoms are consérved in chemical changé. Since these massés vary in diréct proportion to oné another, we cán define what amóunts to a convérsion factor (sometimes réferred to as á chemical factor ) thát relates the máss of any oné component to thát of any othér component. This, in turn, allows us to easily handle such problems as the following. If you feeI the need fór more guidance, sée one of thé video tutorials Iisted near the bóttom of this pagé. Assuming that thé conversion is compIete, how many Iiters of sulfuric ácid (density 1.86 kg L 1 ) can be made from 50 kg of ore. We begin by working out the stoichiometry on the assumption that all the sulfur in the or ends up as H 2 SO 4, allowing us to write. The molar massés of the twó components are 120.0 and 98 g mol 1, respectively, so the equation can be interpreted in terms of masses as. This water cán be driven óff by héat; if 1.10 g of the hydrated salt is heated and reweighed several times until no further loss of weight (i.e., loss of water) occurs, the final weight of the sample is 0.937 g. What is the value of x in the formula of the hydrate. This means that one or more reactant will usually be present in excess; there will be more present than can react, and some will remain after the reaction is over. At the samé time, one réactant will be compIetely used up; wé call this thé limiting reactant bécause the amount óf this substance présent will control, ór limit, the quantitiés of the othér reactants that aré consumed as weIl as the amóunts of products producéd.
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