A student neutralises dilute sulfuric acid, H2SO4\text{H}_2\text{SO}_4H2SO4, with potassium hydroxide, KOH\text{KOH}KOH, to make potassium sulfate, K2SO4\text{K}_2\text{SO}_4K2SO4.
This is the equation for the reaction: 2KOH+H2SO4→K2SO4+2H2O2\text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}2KOH+H2SO4→K2SO4+2H2O
The student makes 6.50 g6.50\text{ g}6.50 g of potassium sulfate. The percentage yield is 75%75\%75%.
Calculate the mass of potassium hydroxide the student used in the reaction.
Give your answer to 3 significant figures.
Relative atomic mass (ArA_rAr): H=1.0\text{H} = 1.0H=1.0 \quad O=16.0\text{O} = 16.0O=16.0 \quad S=32.1\text{S} = 32.1S=32.1 \quad K=39.1\text{K} = 39.1K=39.1