How many grams of CaSO4 (molar mass = 136.2 g/mol) can be produced from the reaction of 0.0126 mol CaBr2 with excess Li2SO4? What is the percent yield if 1.56g CaSO4, is produced?
CaBr2 + Li2SO4 --> CaSO4+ 2LiBr

The real answer is keeping the world healthy so we won’t die

Ithink the answer is d

Use the universal gas law formula: pv = nrt now, the key word in this problem is stp, so at stp, temperature will always be 0 degrees celsius and pressure will always be 1.0 atm or 760 torr. given something is at stp, we already know the two variables pressure and temperature. the other thing we can find with the given information is moles. to find moles, just divide grams by molar mass of the element. 16.0g / 2(16.00)g/mol = 0.5 moles plug the values in to the equation pv = nrt (1.0 atm)( x ) = (0.5 )(0+273=273k) the r value is just a constant for when we use atm, liters, moles and kelvin. i'm assuming you want your answer in liters, but if you dont just move the decimal 3 to the right in your final answer. so when plugging these numbers in to your calculator and trying to find x, you get 11.20665 liters using significant figures the answer is 11.2 liters

5555555555555555555555555555555555555555