Answer:
a. 92.4%
Explanation:
Based on the reaction:
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)
To obtain the percent yield you need to obtain moles of trona and calculate thoeretical moles of Na₂CO₃, and the ratio of obtained moles / theoretical moles of Na₂CO₃ give percent yield, thus:
Moles of trona:
1.00 metric ton × (1x10³kg / 1 metric ton) × ( 1000moles /226.03 kg) = <em>4424 moles</em>
The theoretical moles of Na₂CO₃ that produce 4424 moles of trona are (Based on the reaction, 2 moles of trona produce 3 moles of Na₂CO₃):
4424 moles trona × (3 moles Na₂CO₃ / 2 moles trona) = <em>6636 moles of Na₂CO₃.</em>
The obtained moles of Na₂CO₃:
0.650 metric ton × (1x10³kg / 1 metric ton) × (1000 moles / 105.99kg) = <em>6133 moles</em>
The ratio of obtained moles / theoretical moles gives:
6133 moles / 6636 moles = 0.924 = <em>92.4%</em>
I hope it helps!
They react to form salt ZnCl + hydrogen gas
Answer:
testable questions are answer through observation or an experiment that provides evidence that the questions connects to the scientific concepts rather the opinion feelings
therefore the question can be tested through observation or experiment
Na + NaNO3 = Na2O + N2
4 Na + 2 NaNO3 = 6 Na2O + N2
6 Na on each side
2 N on each side
6 O on each side
