Moles in 72 grams of NaHCO3
1 answer:
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Answer:
21.8 grams.
Explanation:
Molar mass data from a modern periodic table:
- Mg: 24.301;
- O: 15.999.
How many moles of MgO will be produced if Mg is the limiting reactant?
Number of moles of Mg:
.
The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.
How many moles of MgO will be produced if O₂ is the limiting reactant?
Number of moles of O₂:
.
The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. of MgO will be produced if O₂ is in excess.
How many moles of MgO will be produced?
0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.
What's the mass of 0.541284 moles of MgO?
Formula mass of MgO:
.
Mass of 0.541284 moles of MgO:
.
Charles law gives the relationship between temperature of gas and volume of gas.
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,
V = 251 mL
the new volume is 251 mL
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,
V = 251 mL
the new volume is 251 mL