In order to find out the %mass dolomite in the soil,
calculate for the mass of dolomite using the information given from the
titration procedure. You would need to multiply 57.85 ml with 0.3315 M HCl and
you would get the amount of HCl in millimoles. Then multiply the amount of HCl
with 1/2 (given that for every 1 mol of dolomite, 2 mol of HCl would be
needed). Convert the amount of dolomite to mass by multiplying the millimoles
with the molecular weight which is 184.399. Then convert the mass to grams
which is 1.768 grams. Divide the mass of dolomite (1.768 grams) with the weight
of soil sample. The % mass is 7.17.
a. 301 cg
b. 6.2 km
Explanation:
a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:
if 1 g is equal to 100 cg
then 3.01 g are equal to X cg
X = (3.01 × 100) / 1 = 301 cg
b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:
if 1 km is equal to 1000 m
then Y km are equal to 6200 m
Y = (6200 × 1) / 1000 = 6.2 km
Learn more about:
converting units of measurement
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Answer:
a. NH3 is limiting reactant.
b. 44g of NO
c. 40g of H2O
Explanation:
Based on the reaction:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)
4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.
To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:
<em>a. </em><em>Moles NH3 -Molar mass. 17.031g/mol-</em>
25g NH3*(1mol/17.031g) = 1.47moles NH3
Moles O2 = 4 moles
For a complete reaction of 4 moles of O2 are required:
4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.
As there are just 1.47 moles, NH3 is limiting reactant
b. Moles NO:
1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO
Mass NO -Molar mass: 30.01g/mol-
1.47mol NO * (30.01g/mol) = 44g of NO
c. Moles H2O:
1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O
Mass H2O -Molar mass: 18.01g/mol-
2.205mol H2O * (18.01g/mol) = 40g of H2O
The mass of NaCl needed for the reaction is 91.61 g
We'll begin by calculating the number of mole of F₂ that reacted.
- Gas constant (R) = 0.0821 atm.L/Kmol
PV = nRT
1.5 × 12 = n × 0.0821 × 280
18 = n × 22.988
Divide both side by 22.988
n = 18 / 22.988
n = 0.783 mole
Next, we shall determine the mole of NaCl needed for the reaction.
F₂ + 2NaCl —> Cl₂ + 2NaF
From the balanced equation above,
1 mole of F₂ reacted with 2 moles of NaCl.
Therefore,
0.783 mole F₂ will react with = 0.783 × 2 = 1.566 moles of NaCl.
Finally, we shall determine the mass of 1.566 moles of NaCl.
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass = mole × molar mass
Mass of NaCl = 1.566 × 58.5
Mass of NaCl = 91.61 g
Therefore, the mass of NaCl needed for the reaction is 91.61 g
Learn more about stiochoimetry: brainly.com/question/25830314
