How much heat is required to decompose 25.5 grams of NaHCO3? 2NaHCO3(s) + 129 kJ 2Na2CO3(s) + H2O(g) + CO2(g)
how many moles of sodium bicarbonate is in 25.5 grams>
do it as a proportion:
2moles/128kJ=molesbicarbonate/heat and solve for heat
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How much heat is required to decompose 25.5 grams of NaHCO3? 2NaHCO3(s) + 129 kJ 2Na2CO3(s) + H2O(g) + CO2(
To calculate the amount of heat required to decompose 25.5 grams of NaHCO3, we need to use the equation given:
2NaHCO3(s) + 129 kJ → 2Na2CO3(s) + H2O(g) + CO2(g)
First, we need to determine the molar mass of NaHCO3, which is:
Na = 23 g/mol
H = 1 g/mol
C = 12 g/mol
O = 16 g/mol
So, the molar mass of NaHCO3 is:
23 + 1 + 12 + (3 * 16) = 84 g/mol
Next, we can calculate the number of moles of NaHCO3 using the given mass and molar mass:
25.5 g / 84 g/mol ≈ 0.304 moles
Now that we have the number of moles of NaHCO3, we can use the balanced equation to determine the heat required to decompose it. From the equation, we can see that 2 moles of NaHCO3 require 129 kJ of heat. Therefore, 0.304 moles of NaHCO3 will require:
(0.304 moles / 2 moles) * 129 kJ ≈ 19.7 kJ
Therefore, approximately 19.7 kJ of heat is required to decompose 25.5 grams of NaHCO3.