Answer:
25.0 g of sodium carbonate are present in 220 ml of the solution.
Explanation:
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<em>Calculate the volume in milliliters of a 1.07 M sodium carbonate solution that contains 25.0g of sodium carbonate Na2CO3. Be sure your answer has the correct number of significant digits.</em>
First, let's find how many moles of sodium carbonate have a mass of 25.0 g.
The molar mass of Na₂CO₃ is 106 g/mol.
So, if 106 g of sodium carbonate is equal to 1 mol, then 25.0 g will be:
25.0 g · (1 mol / 106 g) = 0.236 mol Na₂CO₃
The solution contains 1.07 mol sodium carbonate in 1 liter.
So, if 1.07 mol sodium carbonate is present in 1 l solution, then, 0.236 mol will be present in:
0.236 mol · (1000 ml / 1.07 mo) = 221 ml (220 ml without any intermediate rounding).
25.0 g of sodium carbonate are present in 220 ml of the solution.
The theoretical yield is calculated as follows
write the equation for reaction
that is 2 Zn + I2 ------> 2 znI
then find the moles of Zn used
moles= mass/molar mass
= 125/ 65.4 = 1.911moles
by use of mole ratio between Zn to Zni which is 2:2 the moles of ZnI is also= 1.911moles
theoretical yield is therefore= moles of ZnI x molar mass of znI
= 1.911 moles x 319.2 g/mol = 609.99 grams