How many grams of water react to form 6.21 moles of Ca (OH) 2

The balanced equation for the formation of Ca (OH) 2 is as follows:

Ca + 2 H2O ... > Ca (OH) 2 + H 2

From the balanced equation, we can note that:

2 moles of water are required to produce 1 mole of Ca (OH) 2, therefore, the number of moles of water needed to produce 6.21 moles of Ca (OH) 2 can be calculated using cross multiplication as follows:

moles of water = (6.21*2) / 1 = 12.42 moles

The molar mass of water is known to be 18 grams

number of moles = mass / molar mass

mass = molar mass * number of moles

mass = 18 * 12.42 = 223.56 grams of water

224 g Since you didn't actually tell us what you're reacting with the water that produces Ca (OH) 2, I'm going to make an educated guess and assume that you're reacting water with Calcium carbide which will produce Ca (OH) 2 and C2H2. If that's the case, then the balanced formula is CaC2 + 2H2O = > Ca (OH) 2 + C2H2 Looking at the balanced formula, it tells use that for every mole of Ca (OH) 2 produced, it takes 2 moles of H2O. So we need: 2 * 6.21 mol = 12.42 mol Now we need to calculate the molar mass of H2O by adding up the atomic weights of all the atoms, so 2 * 1.00794 + 15.999 = 18.01488 g/mol Finally, multiply the molar mass by the number of moles needed to get the answer: 18.01488 g/mol * 12.42 mol = 223.7448096 g Rounding the result to 3 significant figures gives us 224 g