Question

Strontium nitrate, Sr(NO3)2, is used in fireworks to produce brilliant red colors, Suppose we need to prepare 366.6 ml. of 0.115 M SH(NO3)2 solution. How many grams of strontium

nitrate are required?
m(Sr(NO3)2) =

Answer 1

`\boxed{\text{8.91 g}}`

Step-by-step explanation:

1. Calculate the moles of Sr(NO₃)₂

`n = \text{366 mL} * \frac{\text{0.115 mmol}}{\text{1 mL}}= \text{42.09 mmol}`

2. Calculate the mass of SrNO₃)₂

`m = \text{42.09 mmol} * \frac{\text{211.63 mg}}{\text{1 mol}}= \text{8910 mg} = \text{8.91 g}\\\text{The mass of strontium nitrate required is }\boxed{\textbf{8.91 g}}`