How many grams of ammonium carbonate are needed to prepare 1.75 liters of a 0.33

Question

How many grams of ammonium carbonate are needed to prepare 1.75 liters of a 0.333 M solution. A compound is found to contain 30.51 % carbon, 1.69% hydrogen, and 67.80% oxygen by mass and has a molecular mass of 236 g/mole. What is the molecular formula for the compound? Calcium hydroxide, 50 mL, is titrated with 125 mL of 0.20M sulfuric acid. What is the molarity of the sodium hydroxide? Butane gas is used as a fuel for camp stoves, producing heat as a byproduct of the reaction. When 4.14 grams of butane reacts with 12.50 grams of oxygen, the reaction produces 11.0 grams of carbon dioxide. What is the percent yield of the reaction? Cisplatin, Pt(NH_3)_2 Cl_2, is a chemotherapeutic agent that disrupts the growth of DNA. If the current cost of Pt is $1118.00 per troy ounce (1 troy oz = 31.10 grams), how many grams of cisplatin can you make with $3000.00 worth of platinum?

Explanation / Answer

Q16.

g of (NH4)2CO3

MW of (NH4)2CO3 = 96.09 g/mol

for

V = 1.75 L of M = 0.333 M

first, find moles

mol = MV = 1.75*0.333 = 0.58275 mol

mass = mol*MW =0.58275*96.09 = 55.996 g of salt rquired

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