NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
<span>Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27% </span>
<span>3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
</span><span>0.8211 grams Na + 1.266 grams Cl = 2.087 grams</span>
Fe^2 O^3 + 6HCl --> 2FeCl^3 + 3H^2 O
Answer:
55.18 L
Explanation:
First we convert 113.4 g of NO₂ into moles, using its molar mass:
- 113.4 g ÷ 46 g/mol = 2.465 mol
Then we<u> use the PV=nRT formula</u>, where:
- P = 1atm & T = 273K (This means STP)
- R = 0.082 atm·L·mol⁻¹·K⁻¹
Input the data:
- 1 atm * V = 2.465 mol * 0.082atm·L·mol⁻¹·K⁻¹ * 273 K
And <u>solve for V</u>:
I believe the ratio is 1:2
