K2Cr2O7 + 6NaI + 7H2SO4 --> Cr2(SO4)3 + 3I2 + 7H2O + 3Na2SO4 + K2SO4
Part1: You want to produce 1g of I2. The molecular mass of I2 is 2*126.90447g/mol = 253.80894/mol
1g I2 * (1mol / 253.80894g) = 0.00393997154mol I2
Now, using the balanced chemical equation, you can convert moles of I2 to moles of K2Cr2O7
0.00393997154mol I2 * (1mol K2Cr2O7 / 3mol I2) = 0.00131332385mol K2Cr2O7
Now, convert the moles of K2Cr2O7 back to mass, using its molecular mass 294.1846 g/mol.
0.00131332385mol K2Cr2O7 * (294.1846 g/mol) = 0.386359651g K2Cr2O4.
Part2: I'm going to assume the chemical reaction is Na + H2O --> NaOH + H2. Balancing this gives
2Na + 2H2O --> 2NaOH + H2
Since you have excess water, sodium is the limiting reagent. Its molecular mass is 22.99g/mol.
25g Na * (1mol / 22.99g) = 1.08742932mol Na.
Converting this to moles of Hydrogen gas using the equation, we get
1.08742932mol Na * (1mol H2 / 2 mol Na) = 0.54371466mol H2.
This is the theoretical yield.
Keeping this in mind for later, let's get the actual yield. Use the following equation, using the constant R=0.0821 and keeping in mind that NTP (Normal Temperature and Pressure) is equal to 293K and 1atm, respectively:
(1)(15.5) = n(0.0821)(293)
n = 15.5 / (0.0821 * 293) = 0.644348647mol H2
So, theoretical percentage yield refers to actual yield over theoretical yield.
If we crunch the numbers, TPY = (0.644348647/0.54371466) * 100% = 118.508603% yield.
It is strange that the percent yield is over 100%. I could have made an error, but I did my best to check my work.