Reaction Stoichiometry Lab Report

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Molar Mass The mass in grams of 1 mole units of a substance. Denoted by g/mol. The molar mass of one mole of atoms of an element is equal to the atomic mass of the element expressed in g/mol, as said earlier, “amu” can be replaced with “g/mol”. Section III. Reaction Stoichiometry As said in the introduction, reaction stoichiometry allows us to determine the amount of substance consumed in a chemical reaction. It is also the study of the quantitative relationship of the reactants and products in a balanced equation. A balanced equation is needed because of the Law of conservation of mass, which states that matter cannot be created nor destroyed. The Concept of Reaction Stoichiometry To produce ammonia, nitrogen gas should react with …show more content…

Mole to Mole Calculation We use this method if we want to know much of a substance is consumed (or produced) in a chemical reaction (reactant or product). Problem 3.1. Mole to mole calculation How many moles of nitrogen gas (N2) are consumed to produce 3.00 mol of ammonia (NH3)? Answer: To do this, we follow these steps. Step 1: Always write first the balanced chemical equation. N_2(g) + 〖3H〗_(2(g)) → 〖2NH〗_3(g) Step 2: We write the stoichiometric relation (or equivalence) between the known substance and the unknown substance from the balanced chemical equation. We write the mole ratio (conversion factor with dimensional analysis): (Unknown subtance)/(Known substance)=(1 mol N_2)/(2 mol 〖NH〗_3 ) Step 3: Multiplication We multiply the known substance (number of moles) by the conversion factor (mole ratio) to obtain the unknown substance (number of …show more content…

Answer Follow the procedure in figure 3.2. Step 1: Always write first the balanced chemical equation. SiH_(4(g))→〖Si〗_((g))+2H_(2(g)) Step 2: Follow mass A → mol A relationship given in figure 3.2. Mass Si →⌊(mol Si)/(mass Si)⌋→ Mole Si Mol Si = 0.22g Si x (1 mol Si)/(28.00g Si) = 7.86 x 〖10〗^(-3) mol Si Step 3: We write the stoichiometric equivalence (also ratio) between the known substance and the unknown substance, converting mole of Si to the mole of 〖SiH〗_4. This can be done with the use of mol A→ mol B relationship as depicted in figure 3.2. Mass of SiH_4 = 7.86 x 〖10〗^(-3) mol Si x (1 mol SiH4 )/(1 mol Si) = 7.86 x 〖10〗^(-3) mol SiH_4 Step 4: Convert the number of moles of SiH_4 to its mass in grams using the mol B → mass B relationship in figure 3.2. Mass of SiH_4 = 0.220g Si x (1 mol Si)/(28.00g Si) x (1 mol 〖SiH〗_4)/(1 mol Si) x (32.00g 〖SiH〗_4)/(1 mol 〖SiH〗_4 ) We can now conclude that the mass in grams of 〖SiH〗_4 needed to prepare 0.220g Si is 0.251g 〖SiH〗_4. Other conversions can be made from figure 3.2 such as the mass to mole calculation. To do that we only need to cut the equation from mass A up to mole B

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