Answer:
The answer is 6.25g.
Explanation:
First create your balanced equation. This will give you the stoich ratios needed to answer the question:
2C8H18 + 25O2 → 16CO2 + 18H2O
Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:
7.72 g / 16 g/mol = 0.482 mol
Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:
x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2
x = 0.347 mol H2O
The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:
0.347 mol x 18 g/mol = 6.25g
2.7 L in cm³ :
2.7 * 1000 = 2700 cm³
Weight = Volume * Density
2700 * 0.79 = 2133 g
1 Ibs = <span>453.59 g
2133 / 453.59 = 4.70 Ibs
hope this helps!</span>
Answer:
Mass of NaBr produced = 23.67 g
Explanation:
Given data:
Mass of AgBr = 42.7 g
Mass of NaBr produced = ?
Solution:
Chemical equation:
2Na₂S₂O₃ + AgBr → NaBr + Na₃(Ag(S₂O₃)₂
Number of moles of AgBr:
Number of moles = mass/molar mass
Number of moles = 42.7 g/ 187.7 g/mol
Number of moles = 0.23 mol
now we will compare the moles of AgBr with NaBr.
AgBr : NaBr
1 : 1
0.23 : 0.23
Mass of NaBr:
Mass = number of moles × molar mass
Mass = 0.23 mol × 102.89 g/mol
Mass = 23.67 g
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Answer:
If you are given a chemical equation and specific amounts for each reactant in grams, you have to follow these steps, in order, to determine how much product can possilby be made:
1. Convert each reactant into moles of the product.
2. Determine which reactant is the limiting reactant.
3. Convert the moles of product, from the limiting reactant, to grams.
Explanation:
