Answer:
The number of moles of CaCO3 on the bag is 112.90 moles
Explanation:
number mole (n) = mass (m) divided by molecular mass (Mm)
Mm of CaCO3 = 100.0869 g/mole
mass in grams = 11.3 Kg x (10^3 g/1 Kg) = 11300 grams
number of moles (n) = 11300 grams divided by 100.0869 grams per mole = 112.90 moles of CaCO3 in the bag.
Answer:
188.5g of dextrose are needed
Explanation:
In Weight per volume percentage - %(w/v) -, the concentration is defined as the mass of solute in grams -In this case, dextrose-, in 100mL of solution.
As you want to prepare 725mL of a 26.0% (w/v) solution. you need:
725mL * (26g / 100mL) = 188.5g of solute =
<h3>188.5g of dextrose are needed</h3>
Answer:
The mass of 2.35 moles of (NH4)3PO4 is 411.156 or 411.156 grams/g.
Explanation:
By calculating the mass of the formula, (NH4)3PO4, which equals up to 174.96 grams you divide 2.35 moles by 174.96 grams to equal 411.156 grams. The way to find out the mass is to multiple the molar mass of each chemical to the number beside it such as H multiplied by 4, P multiplied by 3, and so on.
<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>
Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.
Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3
To find for the theoretical yield, we first determine the limiting reactant.
100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2
Therefore, the limiting reactant is O2.
Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3
Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%