Calculate The Number Of Moles In 17.8 G Of The Antacid Magnesium Hydroxide, Mg(oh)2.

The moles of Mg 2+ ions in solution equal the moles of Mg(OH) 2 that dissolved, but the moles of OH – ions in solution are two times the moles of Mg(OH) 2 that dissolved. You can use these relationships to write the solubility product constant expression in terms of one unknown.

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Calculate the number of moles in 17.8 g of the antacid magnesium hydroxide, Mg(OH)2. 4. Calculate the number of oxygen atoms in 29.34 g of sodium sulfate, Na2SO4. 5. Calculate the mass in grams of 8.35 1022 molecules of CBr4. 6. Household sugar, sucrose, has the molecular formula C12H22O11. What is the % of carbon in sucrose, by mass? 7. Terephthalic acid, used in the production of polyester.

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A common antacid contains magnesium hydroxide, Mg(OH)2, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10 M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10 g of Mg(OH)2? Plan We know the mass of Mg(OH)2 (0.10 g) that reacts.

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(c) The number 0.0023 can be factored into 2.3 ϫ 0.001. Since , the number can be expressed as Since , the number can be expressed as Or, we can move the decimal point three spaces to the right.

= 2.12 g H2 2 mol HCl 2.10 mol HCl 1 mol H2 1 mol H2 2.02 g H2 example 1.00 g Zn reacts with 6.2 x 10-3 mol Pb(NO3)2 to form Zn(NO3)2 and Pb… 1) Which is LR? 2) How much Pb will be formed? first, the balanced equation!. example cont’d Zn + Pb(NO3)2 Pb + Zn(NO3)2 find the mols of both players: 1.00 g Zn 0.0153 mol Zn (done already!) 6.2 x 10-3 mol Pb(NO3)2 example cont’d again, pretend.

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If you know the concentration and volume of the silver nitrate solution, you can calculate the number of moles of silver in solution. Sodium chloride will react in a 1:1 ratio with the silver ions to give silver chloride. Multiply the number of moles of silver nitrate by 58.44, the molecular weight of sodium chloride. Then multiply this number by 1.5, so that you have an excess of sodium.

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Thus, the combined mass of the reactants (3.25 g + 3.32 g = 6.57 g) is exactly equal to the combined mass of the products (4.55 g + 2.02 g = 6.57 g). The combined masses of these two reactants.. The combined masses of these two reactants..

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